MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
ELEMENTARY PROBLEMS ON PERMUTATION & COMBINATION NOTE: USE FUNDAMENTAL PRINCIPLE OF COUNTING & ENJOY DOING THE FOLLOWING.
Q.11/1 In how many ways can clean & clouded (overcast) days occur in a week assuming that an entire day is
either clean or clouded. [ Ans. 27 = 128 ]
Q.22/1 Four visitors A, B, C & D arrive at a town which has 5 hotels. In how many ways can they disperse themselves among 5 hotels, if 4 hotels are used to accommodate them. [ Ans. 5 . 4 . 3 . 2 = 120] Q.33/1 If the letters of the word “VARUN” are written in all possible ways and then are arranged as in a
dictionary, then the rank of the word VARUN is :
(A) 98 (B) 99 (C*) 100 (D) 101
Q.44/1 How many natural numbers are their from 1 to 1000 which have none of their digits repeated. [Hint: S D = 9 ; DD = 9 · 9 = 81 ; TD = 9 · 9 · 8 = 648] [ Ans. 738] Q.5 3 different railway passes are allotted to 5 students. The number of ways this can be done is :
(A*) 60 (B) 20 (C) 15 (D) 10
Q.66/1 There are 6 roads between A & B and 4 roads between B & C. (i) In how many ways can one drive from A to C by way of B ?
(ii) In how many ways can one drive from A to C and back to A, passing through B on both trips ? (iii) In how many ways can one drive the circular trip described in (ii) without using the same road more
than once. [ Ans. (i) 24 ; (ii) 576 ; (iii) 360 ]
Q.77/1 (i) How many car number plates can be made if each plate contains 2 different letters of English alphabet, followed by 3 different digits.
(ii) Solve the problem, if the first digit cannot be 0. (Do not simplify)
[ Ans. (i) 26 . 25 . 10 . 9 . 8 = 468000 ; (ii) 26.25.9.9.8 = 421200 ] Q.88/1 (i) Find the number of four letter word that can be formed from the letters of the word HISTORY.
(each letter to be used at most once) [ Ans. 7 . 6 . 5 . 4 = 42 x 20 = 840 ] (ii) How many of them contain only consonants? [ Ans. 5 . 4 . 3 . 2 = 120 ] (iii) How many of them begin & end in a consonant? [ Ans. 5 . 4 . 4 . 4 = 400 ] (iv) How many of them begin with a vowel? [ Ans. 240 ] (v) How many contain the letters Y? [ Ans. 480 ] (vi) How many begin with T & end in a vowel? [ Ans. 40 ] (vii) How many begin with T & also contain S? [ Ans. 60 ] (viii) How many contain both vowels? [ Ans. 240 ] Q.99/1 If repetitions are not permitted
(i) How many 3 digit numbers can be formed from the six digits 2, 3, 5, 6, 7 & 9 ? [ Ans. 120 ]
(ii) How many of these are less than 400 ? [ Ans. 40 ]
(iii) How many are even ? [ Ans. 40 ]
(iv) How many are odd ? [ Ans. 80 ]
(v) How many are multiples of 5 ? [ Ans. 20 ]
Q.10 How many two digit numbers are there in which the tens digit and the units digit are different and odd? [ Ans. 5 . 4 = 20 ]
Q.1111/1 Every telephone number consists of 7 digits. How many telephone numbers are there which do not 7 CLASS : XIII (VXYZ) Special DPP on Permutation and Combination DPP. NO.-1
[2] Q.1212/1 (a) In how many ways can four passengers be accommodate in three railway carriages, if each
carriage can accommodate any number of passengers.
(b) In how many ways four persons can be accommodated in 3 different chairs if each person can occupy only one chair. [ Ans. (a) 34 ; (b) 4.3.2 = 24 ]
Q.13 How many odd numbers of five distinct digits can be formed with the digits 0,1,2,3,4 ? [ Hint: 3 × 3 × 2 × 1 × 2 = 36 Ans ]
Q.1414/1 Number of natural numbers between 100 and 1000 such that at least one of their digits is 7, is
(A) 225 (B) 243 (C*) 252 (D) none
[Hint: 9 · 10 · 10 = 900 [ 13th Test (5-12-2004)]
Total no. of numbers without 7 = 81 × 8 = 648
required number = 900 – 648 = 252 ]
Q.1515/1 How many four digit numbers are there which are divisible by 2. [Ans. = 4500] Q.16 The 120 permutations of MAHES are arranged in dictionary order, as if each were an ordinary five-letter
word. The last letter of the 86th word in the list is
(A) A (B) H (C) S (D*) E [12th (14-5-2006)]
[Sol. The first 24 = 4! words begins with A, the next 24 begin with E and the next 24 begin with H. So the 86th begins with M and it is the 86 – 72 = 14th such word. The first 6 words that begin with M begin with MA and the next 6 begin with ME. So the desired word begins with MH and it is the second such word. The first word that begins with MH is MHAES, the second is MHASE. Thus E is the letter we seek. ]
Q.1717/1 Find the number of 7 lettered palindromes which can be formed using the letters from the English
alphabets. [Ans. 264]
[Hint: A palindrome is a word or a phase that is the same whether you read is forward or backword. e.g. REFER]
Q.1818/1 Number of ways in which 7 different colours in a rainbow can be arranged if green is always in the
middle. [ Ans 720 ]
Q.1919/1 Two cards are drawn one at a time & without replacement from a pack of 52 cards. Determine the number of ways in which the two cards can be drawn in a definite order. [Ans. 52 × 51= 2652]
Q.20 Find the number of ways in which the letters of the word "MIRACLE" can be arranged if vowels always occupy the odd places.
Hint: [ Ans: 4·3·2·4·3·2·1=24·24=576 ]
Q.2121/1 Numbers of words which can be formed using all the letters of the word "AKSHI", if each word begins with vowel or terminates in vowel. [ Ans: 2·24+2·24-2·6 = 84 ]
Q.2222/1 A letter lock consists of three rings each marked with 10 different letters. Find the number of ways in which it is possible to make an unsuccessful attempts to open the lock. [Ans: 103 - 1 = 999] Q.2323/1 How many 10 digit numbers can be made with odd digits so that no two consecutive digits are same.
[Ans: 5·49] Q.24 In how many ways can the letters of the word "CINEMA" be arranged so that the order of vowels do
not change. [Ans: 6
3 !
! = 120]
Q.2525/1 How many natural numbers are there with the property that they can be expressed as the sum of the cubes of two natural numbers in two different ways . [Ans. Infinitely many]
MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
Q.11/2 How many of the 900 three digit numbers have at least one even digit?
(A*) 775 (B) 875 (C) 450 (D) 750
[Sol. There are 900 three digit numbers and there are five odd digits. Thus, there are 53 = 125 three digit numbers comprised of only odd digits. The other 900 – 125 = 775 three digit numbers must contain at
least one even digit. ] [12 & 13 05-3-2006]
Q.22/2 The number of natual numbers from 1000 to 9999 (both inclusive) that do not have all 4 different digits is
(A) 4048 (B*) 4464 (C) 4518 (D) 4536 OR
What can you say about the number of even numbers under the same constraints?
[Hint: Total – All four different = 91039·9·8·7 OR
Ans. 2204 ; all 4 digit even number – number of 4 digit even numbers with different digit ]
Q.33/2 The number of different seven digit numbers that can be written using only three digits 1, 2 & 3 under the condition that the digit 2 occurs exactly twice in each number is
(A*) 672 (B) 640 (C) 512 (D) none
[Hint: Two blocks for filling 2 can be selected in 7C2 ways and the digit 2
5 2 7 2 · C can be filled only in one way other 5 blocks can be filled in 25 ways.]
Q.44/2 Out of seven consonants and four vowels, the number of words of six letters, formed by taking four consonants and two vowels is (Assume that each ordered group of letter is a word):
(A) 210 (B) 462 (C*) 151200 (D) 332640 [Hint : 7C
4 4C
2 . 6! = 151200 ]
Q.55/2 All possible three digits even numbers which can be formed with the condition that if 5 is one of the digit, then 7 is the next digit is :
(A) 5 (B) 325 (C) 345 (D*) 365
[Hint : 5 + 8.9.5 = 365; 1st place is five + where 1st place is not five 5 7 | 5 + 5 | 9 | 8 | ]
Q.66/2 For some natural N, the number of positive integral 'x' satisfying the equation , 1 ! + 2 ! + 3 ! + ... + (x !) = (N)2 is :
(A) none (B) one (C*) two (D) infinite
[Hint: x = 1 & x = 3 ]
Q.77/2 The number of six digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 & 7 so that digits do not repeat and the terminal digits are even is :
(A) 144 (B) 72 (C) 288 (D*) 720 [Hint: 1 . . 3 . . 5 . . 7 3C 2 · 2! · 5C 4 · 4! = 6 × 120 = 720 ]
[4] Q.8 In a certain strange language, words are written with letters from the following six-letter alphabet :
A, G, K, N, R, U. Each word consists of six letters and none of the letters repeat. Each combination of these six letters is a word in this language. The word "KANGUR" remains in the dictionary at,
(A*) 248th (B) 247th (C) 246th (D) 253rd [Sol. beginning with A or G = 240 [18-12-2005, 13th]
beginning with 6 247th K A N G R U
248th K A N G U R ]
Q.9 Consider the five points comprising of the vertices of a square and the intersection point of its diagonals. How many triangles can be formed using these points?
(A) 4 (B) 6 (C*) 8 (D) 10
[Hint: To form a triangle, 3 points out of 5 can be chosen in 5C
3 = 10 ways. But of these, the three points lying on the 2 diagonals will be collinear. So 10 – 2 = 8 triangles can be formed]
Q.1010/2 A 5 digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 & 5 without repetition. The total number of ways this can be done is :
(A) 3125 (B) 600 (C) 240 (D*) 216
[Hint: reject 0 + reject 3 5! + 4 · 4! = 120 + 96 = 216 ]
Q.1111/2 Number of 9 digits numbers divisible by nine using the digits from 0 to 9 if each digit is used atmost once is K · 8!, then K has the value equal to ______. [Ans. K = 17] [Hint: Case-I : reject '0' ; Case-II : reject 9
Case-I: 9!; Case-II: 0, 1, 2, ..., 8; 9! – 8! = 8 · 8!; I + II = (17)8!]
Q.1212/2 Number of natural numbers less than 1000 and divisible by 5 can be formed with the ten digits, each digit not occuring more than once in each number is ______ .
[Hint : single digit = 1 ;
two digit: 5 = 8 + 0 = 9 Total = 17
three digit: 5 = 8 · 8 = 64; 0 = 9 · 8 = 72 Total = 136 hence Single digit + two digit + three digit = 154 Ans.]
Q.1311/3 Number of 3 digit numbers in which the digit at hundreath's place is greater than the other two digit is
(A*) 285 (B) 281 (C) 240 (D) 204
[Sol. When all 3 are distinct
z y x 3 2 1st nd rd 10C
3 · 2 (largest being at the 1st place and the
= 240 remaining two can be arranged in two ways) [11th, 16-11-2008] when y = z
10C
2 · 1 (largest on the 1
st place and remaining two being = 45 equal on the 2nd and 3rd place) e.g. (211, 100)]
Total = 285 Ans. ]
Q.1412/3 Number of permutations of 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time, such that the digit 1 appearing somewhere to the left of 2
3 appearing to the left of 4 and 5 somewhere to the left of 6, is
(e.g. 815723946 would be one such permutation)
Sol. Number of digits are 9 [12th, 28-09-2008] [Dpp-3] select 2 places for the digit 1 and 2 in 9C
2 ways
from the remaining 7 places select any two places for 3 and 4 in 7C 2 ways and from the remaining 5 places select any two for 5 and 6 in 5C
2 ways now, the remaining 3 digits can be filled in 3! ways
Total ways = 9C 2 · 7C 2 · 5C 2 · 3! = 2!·7! ! 9 · 2!·5! ! 7 · 2!·3! ! 5 · 3! = 8 ! 9 = 8 ! 7 · 8 · 9 = 9 · 7! Ans. ]
Q.15 Number of odd integers between 1000 and 8000 which have none of their digits repeated, is
(A) 1014 (B) 810 (C) 690 (D*) 1736
[Sol. Let the last place is 9
1/3/5/7/9
then we have 7 · 8 · 7 = 392 (1st place only 1 to 7 can come) If the last place has 1 | 3 | 5 | 7 then
we have (6) (8) (7) (4) = 1344 [0, 9 or 8 can not come at units place] Total = 392 + 1344 = 1736] [11th, 25-01-2009, P-1]
Q.164/3Find the number of ways in which letters of the word VALEDICTORY be arranged so that the vowels may never be separated.
[ Hint: VLDCTRYY or 8! × 4! = 40320 × 24 = 967680 Ans ]
[‘Valediction means farewell after graduation from a college. Valedictory : to take farewell]
Q.177/3The number of ways in which 5 different books can be distributed among 10 people if each person can get at most one book is :
(A) 252 (B) 105 (C) 510 (D*) 10C
5.5! [Hint: Select 5 boys in 10C5 and distribute 5 books in 5! ways hence 10C5. 5!]
Q.188/2A new flag is to be designed with six vertical strips using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent strips have the same colour is (A*) 12 × 81 (B) 16 × 192 (C) 20 × 125 (D) 24 × 216
[Hint: 1st place can be filled in 4 ways 2nd place can be filled in 3 ways
3rd place can be filled in 3 ways and |||ly 4th, 5th and 6th each can be filled in 3 ways. hence total ways = 4 × 35 = 12 × 81 ]
Q.194/4 5 Indian & 5 American couples meet at a party & shake hands . If no wife shakes hands with her own husband & no Indian wife shakes hands with a male, then the number of hand shakes that takes place in the party is :
(A) 95 (B) 110 (C*) 135 (D) 150
[Hint: 20C2 (50 + 5) = 135 ]
Q.20192/1(10/3) There are 720 permutations of the digits 1, 2, 3, 4, 5, 6. Suppose these permutations are arranged from smallest to largest numerical values, beginning from 1 2 3 4 5 6 and ending with 6 5 4 3 2 1. (a) What number falls on the 124th position?
(b) What is the position of the number 321546? [Ans. (a) 213564, (b) 267th] [Sol. (a) digits 1, 2, 3, 4, 5, 6
starting with 2, 1, 3, 4 number of numbers = 2 123rd
finally 124th is = 213564 (b) N = 321546
number of numbers beginning with 1 = 120 number of numbers beginning with 2 = 120 starting with 31 ... = 24 starting with 3214 ... = 2
finally = 1
MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
Q.15/3 How many numbers between 400 and 1000 (both exclusive) can be made with the digits 2,3,4,5,6,0 if (a) repetition of digits not allowed.
(b) repetition of digits is allowed.
[ Hint: (a) 3×5×4 = 60 ; (b) 3×6×6 = 108 – 1 = 107 ]
Q.29/3 The 9 horizontal and 9 vertical lines on an 8 × 8 chessboard form 'r' rectangles and 's' squares. The ratio
r s
in its lowest terms is
(A) 6 1 (B*) 108 17 (C) 27 4 (D) none
[Sol. no. of squares are [12 & 13th test (29-10-2005)] S = 12 + 22 + 32 + ... + 82 = 6 ) 17 )( 9 ( 8 = 204 no. of rectangles r = 9C 2 · 9C2 = 1296 hence r s = 1296 204 = 324 51 = 108 17 ]
Q.311/3A student has to answer 10 out of 13 questions in an examination . The number of ways in which he can answer if he must answer atleast 3 of the first five questions is :
(A*) 276 (B) 267 (C) 80 (D) 1200
[Hint : 13C
10 number of ways in which he can reject 3 questions from the first five or 13C 10 5C 3 = 286 – 10 = 276 or 5C 3 . 8C 7 + 5C 4 . 8C 6 + 5C 5 . 8C 5 = 276 ] [ Note that 5C 3 . 10C
7 is wrong ( cases repeat]
Q.412/3 The number of three digit numbers having only two consecutive digits identical is :
(A) 153 (B*) 162 (C) 180 (D) 161
[Hint : when two consecutive digits are 11, 22, etc = 9 . 9 = 81 when two consecutive digits are 0 0 = 9
when two consecutive digits are 11, 22, 33, ... = 9 . 8 = 72 Total ]
Q.51/4 A telegraph has x arms & each arm is capable of (x 1) distinct positions, including the position of rest. The total number of signals that can be made is ______ . [ Ans. (x 1)x 1 ]
Q.62/4 The interior angles of a regular polygon measure 150º each . The number of diagonals of the polygon is
(A) 35 (B) 44 (C*) 54 (D) 78
[Hint : exterior angle = 30°
Hence number of sides = 30 360 = 12 number of diagonals = 2 ) 3 12 ( 12 = 54 ]
[Note that sum of all exterior angles of a polygon = 2 and sum of all the interior angles of a polygon =(2n-4) 2]
[8] Q.73/4 Number of different natural numbers which are smaller than two hundred million & using only the digits
1 or 2 is :
(A * ) (3).28 2 (B) (3).28 1 (C) 2 (29 1) (D) none [Hint: Two hundred million = 2 x 108; (21+ 22 +23+24+25+26+27+28) + 28 = 766 ]
Q.85/4 The number of n digit numbers which consists of the digits 1 & 2 only if each digit is to be used atleast once, is equal to 510 then n is equal to:
(A) 7 (B) 8 (C*) 9 (D) 10
[Hint: (2 x 2 x ...2) n times-(when 1 or 2 is there at all the n places] [ Ans. 2n 2 ]
Q.96/4 Number of six digit numbers which have 3 digits even & 3 digits odd, if each digit is to be used atmost
once is ______ . [ Ans. 64800 ]
[ when 0 is included (4C 2 .
5C
3 . 5 . 5!) and when 0 is excluded ( 4C
3 . 5C
3 . 6!) ] [Hint: alternatively, 5C
3 · 5C3 · 6! since all digits 0, 1, 2, ...8, 9 are equally likely at all places
required number = 5 3 5 3 6 10 C C ! · 9 digits
or required number of ways = 5C 3 . 5C 3 . 6 ! – 4C 2 . 5C 3 . (6! – 5!) ]
Q.1016/9 Find the number of 10 digit numbers using the digits 0, 1, 2, ... 9 without repetition. How many of these are divisible by 4.
[Sol. Digit 0, 1, 2,...8, 9
For a number to be divisible by 4 the number formed by last two digits must be divisible by 4 and can be 04, 08, 12, ..., 96 ; Total of such numbers = 24
Out of these 44 and 88 are to be rejected. (as repetition is not allowed) Hence accepted number of cases = 22
Out of these number of cases with '0' always included are 04, 08, 20, 40, 60, 80 (six) no. of such numbers with these as last two digits = 6 · 8! ...(1)
e.g. [ × × × × × × × × 04 ]
no. of other numbers = 16 · 7 · 7! = 14 · 8! ....(2) e.g. [ × × × × × × × × 16 ]
Total number = 6 · 8! + 14 · 8! = (20) · 8! Ans. ]
Q.1114/8 There are counters available in x different colours. The counters are all alike except for the colour. The total number of arrangements consisting of y counters, assuming sufficient number of counters of each colour, if no arrangement consists of all counters of the same colour is :
(A*) xy x (B) xy y (C) yx x (D) yx y
Q.128/418 points are indicated on the perimeter of a triangle ABC (see figure). How many triangles are there with vertices at these points?
(A) 331 (B) 408 (C) 710 (D*) 711 [Hint: 18C 3 – 3· 7C 3 = 816 – 105 = 711 ] [08-01-2005, 12 th]
Q.139/4An English school and a Vernacular school are both under one superintendent. Suppose that the superintendentship, the four teachership of English and Vernacular school each, are vacant, if there be altogether 11 candidates for the appointments, 3 of whom apply exclusively for the superintendentship and 2 exclusively for the appointment in the English school, the number of ways in which the different appointments can be disposed of is :
[Hint : 3C 1 .
4C
2 . 2! . 6! ]
Q.1410/4 A committee of 5 is to be chosen from a group of 9 people. Number of ways in which it can be formed if two particular persons either serve together or not at all and two other particular persons refuse to serve with each other, is
(A*) 41 (B) 36 (C) 47 (D) 76 [Sol. 9 other 5 separated D / C AB AB included 7C
3 – 5C1 = 30 (7C3 denotes any 3 from (CD and 5 others – no. of ways when CD) AB excluded 7C
5 – 5C
3 = 11 is taken and one 3 from remaining five) ——
41 ] [18-12-2005, 12th + 13th]
Q.1511/4 A question paper on mathematics consists of twelve questions divided into three parts A, B and C, each containing four questions. In how many ways can an examinee answer five questions, selecting atleast one from each part.
(A*) 624 (B) 208 (C) 1248 (D) 2304
[Hint: 3 (4C
2 . 4C2 . 4C1) + 3 (4C1 . 4C1 . 4C3) = 432 + 194 = 624 ]
Alternative : Total – [no of ways in which he does not select any question from any one section] 12C
5 – 3 · 8C5 ; Note that 4C1 . 4C1 . 4C1 . 9C2 is wrong think !]
Q.1612/4 If m denotes the number of 5 digit numbers if each successive digits are in their descending order of magnitude and n is the corresponding figure, when the digits are in their ascending order of magnitude then (m – n) has the value
(A) 10C 4 (B*) 9C 5 (C) 10C 3 (D) 9C 3 [Hint: m – n = 10C 5 – 9C5 = 252 – 125 = 126 = 9C5 or 9C4 Alternatively: 9C 5 + 9C4 – 9C5 = 9C4 = 9C5 ]
Q.171/5There are m points on a straight line AB & n points on the line AC none of them being the point A. Triangles are formed with these points as vertices, when
(i) A is excluded (ii) A is included. The ratio of number of triangles in the two cases is:
(A*) m n m n 2 (B) m n m n 2 1 (C) m n 2 2 n m (D) m n m n ( ) ( ) ( ) 1 1 1 [Hint: m C n C m C n C mn n m n m . . . . 2 2 2 2 ]
Q.183/5In a certain algebraical exercise book there are 4 examples on arithmetical progressions, 5 examples on permutation-combination and 6 examples on binomial theorem . Number of ways a teacher can select for his pupils atleast one but not more than 2 examples from each of these sets, is ______.
[ Hint: ( 4C
1 + 4C2) ( 5C1 + 5C2) ( 6C1 + 6C2) ] [ Ans. 3150 ] [Alternatively : add one dummy exercies in each and compute 5C
2 · 6C
2 · 7C
Q.191/9 n r n r n r r n C C C
1 0 1 is equal to : (A) n n n ( ) ( ) 1 2 1 (B) n 1 2 (C) n n( 1) 2 (D*) n 2Q.208/5The number of 5 digit numbers such that the sum of their digits is even is :
(A) 50000 (B*) 45000 (C) 60000 (D) none [Hint: 2 10 9 4 ]
MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
Q.1 Number of ways in which 8 people can be arranged in a line if A and B must be next each other and C
must be somewhere behind D, is equal to
(A) 10080 (B*) 5040 (C) 5050 (D) 10100 [Sol. D C AB [12th 11-05-2008]
AB and 6 other is 7! but A and B can be arranged in 2! ways
Total ways = 7! · 2!
when C is behind D
required number of ways =
! 2 ! 2 · ! 7 = 5040 ways Ans.]
Q.22/5 Number of ways in which 7 green bottles and 8 blue bottles can be arranged in a row if exactly 1 pair of green bottles is side by side, is (Assume all bottles to be alike except for the colour).
(A) 84 (B) 360 (C*) 504 (D) none
[Sol. G G G G G G G / B B B B B B B B [12th & 13th 03-12-2006] one gap out of nine can be taken in 9C
1 ways now green remaining 5
gaps remaining 8
5 gaps for remaining 5 green can be selected in 8C5
Hence Total ways 9C
1 · 8C5 = 9 · 1·2·3 6 · 7 · 8 = 72 · 7 = 504 Ans. Alternatively: 9C
6 × 6 = 504 Ans. (think ! how)
(6 bottle in 6 gaps and the remaining in any one of these 6 gaps)]
Q.34/5 The kindergarten teacher has 25 kids in her class . She takes 5 of them at a time, to zoological garden as often as she can, without taking the same 5 kids more than once. Then the number of visits, the teacher makes to the garden exceeds that of a kid by :
(A) 25C 5 24C 5 (B*) 24C 5 (C) 24C 4 (D) none
[Hint: Number of trips which exceeds when one kid is never included 25C
5 24C4 = 24C5 ]
Q.46/5 Seven different coins are to be divided amongst three persons . If no two of the persons receive the same number of coins but each receives atleast one coin & none is left over, then the number of ways in which the division may be made is
(A) 420 (B*) 630 (C) 710 (D) none
[Hint: 1, 2, 4 groups] [Ans. 7
1 2 4 !
! ! ! × 3 ! ]
Q.57/5 Let there be 9 fixed points on the circumference of a circle. Each of these points is joined to every one of the remaining 8 points by a straight line and the points are so positioned on the circumference that atmost 2 straight lines meet in any interior point of the circle. The number of such interior intersection points is :
(A*) 126 (B) 351 (C) 756 (D) none of these
[Hint : Any interior intersection point corresponds to 4 of the fixed points , namely the 4 end points of the intersecting segments. Conversely, any 4 labled points determine a quadrilateral, the diagonals of which intersect once within the circle . Thus the answer is A ]
[12] Q.610/5The number of ways in which 8 distinguishable apples can be distributed among 3 boys such that every
boy should get atleast 1 apple & atmost 4 apples is K · 7P
3 where K has the value equal to
(A) 14 (B) 66 (C) 44 (D*) 22
Q.711/5A women has 11 close friends. Find the number of ways in which she can invite 5 of them to dinner, if two particular of them are not on speaking terms & will not attend together. [Ans. 378]
[Hint: 11C 5 9C3 = 9C5 + 2 9C4 = 3 9C4 = 378 ] 5 to be selected 9C 3 + 9C 4 + 9C 4 or 11C 5 – 9C 3 ]
Q.812/5A rack has 5 different pairs of shoes. The number of ways in which 4 shoes can be chosen from it , so that there will be no complete pair is :
(A) 1920 (B) 200 (C) 110 (D*) 80 [Hint: 5C 4 . 2 4 or 10 8 6 4 4 . . . ! = 80 ]
Q.93/6 There are 10 seats in a double decker bus, 6 in the lower deck and 4 on the upper deck. Ten passengers board the bus, of them 3 refuse to go to the upper deck and 2 insist on going up. The number of ways in which the passengers can be accommodated is _____. (Assume all seats to be duly numbered)
[ Ans. 4C 2. 2!
6C
3. 3! 5! or 172800]
Q.104/6Find the number of permutations of the word "AUROBIND" in which vowels appear in an alphabetical
order. [Ans. 8C
4 · 4 !] [Hint : A, I, O, U treat them alike. Now find the arrangement of 8 letters in which 4 alike and 4 different = 8
4 ! !]
Q.115/6The greatest possible number of points of intersection of 9 different straight lines & 9 different circles in a plane is
(A) 117 (B) 153 (C*) 270 (D) none
[Hint: 9C
2 . 1 + 9C1 . 9C1 . 2 + 9C2 . 2 = 270 ]
Q.126/6An old man while dialing a 7 digit telephone number remembers that the first four digits consists of one 1's, one 2's and two 3's. He also remembers that the fifth digit is either a 4 or 5 while has no memorising of the sixth digit, he remembers that the seventh digit is 9 minus the sixth digit. Maximum number of distinct trials he has to try to make sure that he dials the correct telephone number, is
(A) 360 (B*) 240 (C) 216 (D) none [Hint: ! 2 ! 4 place fifth for ways 2 place 6 ways 10 th place 7 way 1 th th 7 3 3 2 1 x7 = 9 – x6 x6 can take 0 to 9 = 240 ] [12 & 13th test (09-10-2005)]
Q.137/6If as many more words as possible be formed out of the letters of the word "DOGMATIC" then the number of words in which the relative order of vowels and consonants remain unchanged is ______.
[Hint: 0 A I 3! × 5! – 1 = 719 ]
Q.148/6Number of ways in which 7 people can occupy six seats, 3 seats on each side in a first class railway compartment if two specified persons are to be always included and occupy adjacent seats on the same side, is 5! · (k) then k has the value equal to :
(A) 2 (B) 4 (C*) 8 (D) none
[Hint: including the two specified people, 4 others can be selected in 5C
4 ways . The two adjacent seats can be taken in 4 ways and the two specified people can be arranged in 2! ways, remaining 4 people can be arranged in 4! ways .
Q.159/6Number of ways in which 9 different toys be distributed among 4 children belonging to different age groups in such a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more, is :
(A)
5 8 2 ! (B) 9 2 ! (C*)
9 3 2 3 ! ! ! (D) none[Hint : distribution 2, 2, 2 and 3 to the youngest . Now 3 toys for the youngest can be selected in 9C 3 ways, remaining 6 toys can be divided into three equal groups in
6 2 3 3
! ! . !
way and can be distributed in 3! ways 9C 3 .
6 2 3 ! ! =
9 3 2 3 ! ! ! ]Q.1610/6 In an election three districts are to be canvassed by 2, 3 & 5 men respectively . If 10 men volunteer, the number of ways they can be alloted to the different districts is :
(A*) 10 2 3 5 ! ! ! ! (B) 10 2 5 ! ! ! (C) 10 2 2 5 ! ( !) ! (D) 10 2 2 3 5 ! ( !) ! !
[Hint : number of groups of 2, 3, 5 = 10
2 3 5 !
! ! ! & can be deputed only in one way ]
Q.1711/6 Let Pndenotes the number of ways in which three people can be selected out of 'n' people sitting in a row, if no two of them are consecutive.
If, Pn + 1 Pn = 15 then the value of 'n' is :
(A) 7 (B*) 8 (C) 9 (D) 10 [Hint : Pn = n 2C 3 ; Pn + 1 = n 1C 3 ; Hence n 1C 3 -n 2C 3 = 15 n 2C 3 + n 2C 2 n 2C 3 = 15 or n 2C 2 = 15 n = 8 C ]
Q.18 The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is
(A*) 26 (B) 18 (C) 31 (D) none
[Hint: Draw a Venn diagram with circles named
A : the number of integers from 1 to 100 divisible by 2 = 50 B : the number of integers from 0 to 100 divisible by 3 = 33 C : the number of integers from 0 to 100 divisible by 5 = 20 where the overlap portions are the numbers divisible by both A & B , B & C, etc. (i.e. by 6, 10, 15 & 30).
The total number of integers not divisible is then 26. ]
Use : n (A B C) = n (A) + n (B) + n(C) – n(A B) – n (B C) – n (C A) + n (A B C) ]
Q.19203/1(2/6) In how many different ways a grandfather along with two of his grandsons and four grand daughters can be seated in a line for a photograph so that he is always in the middle and the two grandsons are
never adjacent to each other. [Ans. 528]
[Sol. Total number of ways they can sit = 6! × × × × × ×
no. of ways when the two grandsons are always adjacent = 4 · 2! · 4! = 192 where 4 denotes the no. of adjacent positions
(2!) no. of ways in which two sons can be seated
and 4! no. of ways in which the daughter can be seated in the remaining places. required no. of ways = 720 – 192 = 528 Ans ]
Q.209/5A forecast is to be made of the results of five cricket matches, each of which can be win, a draw or a loss for Indian team. Find
(i) the number of different possible forecasts
(ii) the number of forecasts containing 0, 1, 2, 3, 4 and 5 errors respectively [ Ans: 35 = 243 ; 1, 10, 40, 80, 80, 32 ] [Hint: N(0) = 1 ; N(1) = 2.5C
[15]
MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
Q.14/9 There are six periods in each working day of a school. Number of ways in which 5 subjects can be arranged if each subject is allotted at least one period and no period remains vacant is
(A) 210 (B*) 1800 (C) 360 (D) 3600
[Hint: 6C
2 · 5C1 · 4! = 1800 S1 S2 S3 S4 S5 × × × × ×
note that at least one of the subject has to be repeated [13th Test (5-12-2004)] two periods in which one subject is to be repeated 5C
1 · 4! one subject ]
Q.21/7 There are 10 red balls of different shades & 9 green balls of identical shades. Then the number of arranging them in a row so that no two green balls are together is
(A) (10 !) . 11P
9 (B*) (10 !) . 11C9 (C) 10 ! (D) 10 ! 9 !
Q.32/7 Number of ways in which n distinct objects can be kept into two identical boxes so that no box remains empty, is
[Hint: Consider the boxes to be different for a moment. T1 can be kept in either of the boxes in 2 ways, silmilarly for all other things Total ways = 2n
but this includes when all the things are in B1 or B2 number of ways = 2n 2 Since the boxes are identical 2 2
2
n
= 2n 1 1 ]
Q.43/7 A shelf contains 20 different books of which 4 are in single volume and the others form sets of 8, 5 and 3 volumes respectively. Number of ways in which the books may be arranged on the shelf, if the volumes of each set are together and in their due order is
(A) ! 3 ! 5 ! 8 ! 20 (B) 7! (C*) 8! (D) 7 . 8!
[Hint : Volume of each set may be in due order in two ways, either from left to right or from right to left. Now we have
D1 D2 D3 D4 , = 7! × 2! . 2! . 2! = 8! ]
Q.59/7 In a certain college at the B.Sc. examination, 3 candidates obtained first class honours in each of the following subjects: Physics, Chemistry and Maths, no candidates obtaining honours in more than one subject; Number of ways in which 9 scholarships of different value be awarded to the 9 candidates if due regard is to be paid only to the places obtained by candidates in any one subject is _____. [ Ans: 1680 ] [ Hint : 3 candidates in P; 3 in C and 3 in M. Now 9 scholarships can be divided into three groups in
! 3 . ) ! 3 ( ! 9
3 ways and distributed to P, C, M in 3 ) ! 3 ( ! 9 ways ]
Q.65/7 Number of rectangles in the grid shown which are not squares is
(A*) 160 (B) 162 (C) 170 (D) 185
[Sol. Total = 7C2 · 5C2 = 210 – number of squares [13th, 19-10-2008] number of squares = units 4 units 3 units 2 units 1 3 8 15 24 = 50
required number = 210 – 50 = 160 Ans. ]
Q.76/7 All the five digits number in which each successive digit exceeds its predecessor are arranged in the increasing order of their magnitude. The 97th number in the list does not contain the digit
(A) 4 (B*) 5 (C) 7 (D) 8
[Sol. All the possible number are 9C
5 (none containing the digit 0) = 126 [12
th test (29-10-2005)] Total numbers starting with 1 = 8C
4 = 70 (using 2, 3, 4, 5, 6, 7, 8, 9)
Total starting with 23 = 6C 3 = 20 (4, 5, 6, 7, 8, 9)
Total starting with 245 = 4C 2 = 6 (6, 7, 8, 9)
97th number = ]
Q.87/7 The number of combination of 16 things, 8 of which are alike and the rest different, taken 8 at a time is
______. [Ans. 256] [Hint : A A A A A A A A D1 D2 ... D8 8C 0 + 8C 1 + 8C 2 + ... + 8C 8 = 256 ]
Q.98/7 The number of different ways in which five 'dashes' and eight 'dots' can be arranged, using only seven of these 13 'dashes' & 'dots' is :
(A) 1287 (B) 119 (C*) 120 (D) 1235520 [Hint: 7C 0 + 7C 1 + 7C 2 + 7C 3 + 7C 4 + 7C 5 ]
Q.1010/7 There are n identical red balls & m identical green balls . The number of different linear arrangements consisting of "n red balls but not necessarily all the green balls" is xC
y then (A) x = m + n , y = m (B*) x = m + n + 1 , y = m (C) x = m + n + 1 , y = m + 1 (D) x = m + n , y = n
[Hint: Put one more red ball & find the arrangement of n + 1 red and m green balls = m + n + 1C m ]
Q.116/8A gentleman invites a party of m + n (m n) friends to a dinner & places m at one table T1 and n at another table T2 , the table being round . If not all people shall have the same neighbour in any two arrangement, then the number of ways in which he can arrange the guests, is
(A*) (m n) ! mn 4 (B) 1 2 (m n) ! mn (C) 2 (m n) ! mn (D) none [Hint: 2 )! 1 n ( . 2 )! 1 m ( ! n ! m )! n m ( (A) ]
Q.1211/7 Consider a determinant of order 3 all whose entries are either 0 or 1. Five of these entries are 1 and four of them are '0'. Also aij = aji 1 i, j 3. Find the number of such determinants.
[Ans. 12]
[Sol. We have aij = aji a12 = a21 = a (say) [11th, 28-11-2009, P-2] [Dpp, p&c]
a31 = a13 = b and a23 = a32 c b c a b a 1, 1, 1, 1, 1 and 0, 0, 0, 0 Case-I :
If diagonal elements are (1, 1, 1 )
Conjugate element are, (0, 0), (0, 0), (1, 1) as shown.
1 1 0 1 1 0 0 0 1
This can be done in
! 2
! 3
[17] Case-II :
If diagonal has (1, 0, 0) then the conjugate elements are (1, 1), (1, 1), (0, 0).
Now conjugate elements can be arranged in
2 ! 3 = 3 ways 0 0 1
and diagonal elements can be filled in
2 ! 3
= 3 ways Hence 3 × 3 = 9 ways
Total determinant = 3 + 9 = 12 Ans.]
Q.131/8 Number of different words that can be formed using all the letters of the word "DEEPMALA" if two vowels are together and the other two are also together but separated from the first two is
(A) 960 (B) 1200 (C) 2160 (D*) 1440
[Hint : | D | P | M | L | can be arranged in 4! ways & the two gaps out of 5 gaps can be selected in 5C
2 ways.
two A's and two E's can be arranged in ! 2 ! 2 ! 4 = 6 ways. 4! · 5C 2 · 2!2! ! 4 = 1440 ]
Q.14 A four digit number is called a doublet if any of its digit is the same as only one neighbour. For example, 1221 is a doublet but 1222 is not. Number of such doublets are
(A) 2259 (B*) 2268 (C) 2277 (D) 2349
[Sol. Case-1: abbc [13th, 14-02-2009] [Dpp, P&C] done
We can choose a in 9 ways, b in 9 ways, and c in 9 ways. Hence, there are 93 possibilities for this case
Case-2: bbac
There are 9 possibilities for b, and 9 possibilities for each a and c. Once again, we have that there are 93 possibilities for this case
Case-3: acbb
Once again, there are 9 possibilities for a and c, and 9 possibilities for b. Hence , there are 93 possibilities for this case
Case-4: aabb
We have 9 possibilities for a and 9 possibilities for b. We thus have 81 possibilities for this case. We can now sum up our answers, to get a total of 3 · 93 + 81 = 2268 total ways Ans.
Alternatively: Total four digit numbers = 9000
let us compute the number which are not doublet (a) when no two adjacent digits are same
= 9 · 9 · 9 · 9 = 6561
(b) 3 digits in a row are same at the end of the number if they are non zero then we have
= 9 · 8 = 72
if they are same then = 9
total for (b) = 81
(c) 3 digits at the beginning of the numbers are same, then we have = 9 · 9 = 81
(d) All four digits are same = 9
Q.153/8In a unique hockey series between India & Pakistan, they decide to play on till a team wins 5 matches. The number of ways in which the series can be won by India, if no match ends in a draw is :
(A*) 126 (B) 252 (C) 225 (D) none
[Hint : India wins exactly in 5 matches looses in none 5C 0 ways
India wins exactly in 6 matches wins the 6th and looses anyone in the 1st five 5C 1 ways and so on . 5C 0 + 5C 1 + 6C 2 + 7C 3 + 8C 4 = 126
Alternatively : Total number of ways in which Series can be won by India or Pakistan = 10C 5
required number of ways =
10 5
2 C
= 126 ]
Q.1612/9 Sameer has to make a telephone call to his friend Harish, Unfortunately he does not remember the 7 digit phone number. But he remembers that the first three digits are 635 or 674, the number is odd and there is exactly one 9 in the number. The maximum number of trials that Sameer has to make to be successful is
(A) 10,000 (B*) 3402 (C) 3200 (D) 5000
[Sol. There are 2 ways of filling the first 3 digits, either 635 or 674. Of the remaining 4 digits, one has to be 9 and the last has to be odd. If the last digit is 9 then there are 9 ways of filling each of the remaining 3 digits. thus the total number of phone numbers that can be formed are 2 × 93 = 1458.
If the last digit is not 9, then there are only 4 ways of filling the last digit. (one of 1, 3, 5 and 7). The 9 could occur in any of the 3 remaining places and the remaining 2 places can be filled in 92 ways. Thus the total number of such numbers is : 2 × 4 × 3 × 92 = 1944 1944 + 1458 = 3402 Ans ]
Q.176/9A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. If internal arrangement inside the car does not matter then the number of ways in which they can travel, is
(A) 91 (B) 182 (C*) 126 (D) 3920
[Sol. They can sit in groups of either 5 and 3 or 4 and 4
required number = 1 ! 3 ! 5 ! 8 + 4! 4! 2! ! 2 ! 8 = 126 ] [12th test (05-06-2005)] Q.183/10 One hundred management students who read at least one of the three business magazines are surveyed
to study the readership pattern. It is found that 80 read Business India, 50 read Business world, and 30 read Business Today. Five students read all the three magazines. How many read exactly two magazines?
(A*) 50 (B) 10 (C) 95 (D) 65 [Hint: n (A B C) = 100 n (A) = 80 ; n (B) = 50 ; n (C) = 30 now n (A B C) =
n(A) –
n(A )B + n (A B C) 100 = 80 + 50 + 30 –
n(A )B + 5
n(A )B = 65 now n (E2) =
n(A )B – 3n(A B C) = 65 – 15 = 50]Q.1913/9 Six people are going to sit in a row on a bench. A and B are adjacent, C does not want to sit adjacent to D. E and F can sit anywhere. Number of ways in which these six people can be seated, is
(A) 200 (B*) 144 (C) 120 (D) 56
[Hint: ; C and D separated ; E and F any where [11-12-2005, 11th (PQRS)] and E, F can be seated in 3! 2!
no. of gaps are 4
|
|
E|
F|
C D can be seated in 4C2 · 2!Total ways 3! · 2! · 4C
[19] MATCH THE COLUMN:
Q.2026/10 Column-I Column-II
(A) Number of increasing permutations of m symbols are there from the n set (P) nm numbers {a1, a2, , an} where the order among the numbers is given by
a1 < a2 < a3 < an–1 < an is
(B) There are m men and n monkeys. Number of ways in which every monkey (Q) mC n has a master, if a man can have any number of monkeys
(C) Number of ways in which n red balls and (m – 1) green balls can be arranged (R) nC m in a line, so that no two red balls are together, is
(balls of the same colour are alike)
(D) Number of ways in which 'm' different toys can be distributed in 'n' children (S) mn if every child may receive any number of toys, is
[Ans. (A) R; (B) S; (C) Q; (D) P] [11th pqrs & J (24-11-2006)] [Sol. (A) From n elements select m in nC
m ways and can be arranged only in one way. (B) 1st working can be given to M
1 M2 M3... Mm ® 'm' men any one of m must k1 k2 k3...kn 'n' monkey
|||ly 2nd, 3rd , ..., nth monkey in m way total ways = mn
(C) Number of gaps = m select n gaps in mC
n ways for n red balls total ways = mCn (D) 1st toy in n ways
2nd toy in n ways
MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
Q.14/8 Number of cyphers at the end of 2002C 1001 is (A) 0 (B*) 1 (C) 2 (D) 200 [Hint: 2002C 1001 = (1001)!(1001)! )! 2002 (
no. of zeros in (2002)! are
400 + 80 + 16 + 3 = 499
no. of zeroes in (1001 !)2 = 2(200 + 40 + 8 + 1) = 498
Hence no. of zeroes is 2 ) ! 1001 ( )! 2002 ( = 1 ] [12th (25-9-2005)]
Q.25/8 Three vertices of a convex n sided polygon are selected. If the number of triangles that can be constructed such that none of the sides of the triangle is also the side of the polygon is 30, then the polygon is a (A) Heptagon (B) Octagon (C*) Nonagon (D) Decagon
[Sol. nC
3 – [n + n(n – 4)] = 30
n = 9 satisfies it (C) ] [11th, 13-01-2008]
Q.31/10Given 11 points, of which 5 lie on one circle, other than these 5, no 4 lie on one circle. Then the maximum number of circles that can be drawn so that each contains atleast three of the given points is :
(A) 216 (B*) 156 (C) 172 (D) none
[Hint: 5C
2 . 6C1 + 6C2 5C1 + 6C3 + 1 = 156 alternatively 11C
3 5C3 + 1]
Q.4 Number of 5 digit numbers divisible by 25 that can be formed using only the digits 1, 2, 3, 4, 5 & 0 taken five at a time is
(A) 2 (B) 32 (C*) 42 (D) 52
[ Hint : 1, 2, 3, 4, 5, 0
A number divisible by 25 if the last two digits are 25 or 50 (i) Now, if 5 is not taken then number of numbers = 0
(ii) If 0 is not taken = 6
(iii) if 2 is not taken then = 6 and (iv) If 1 | 3 | 4 is rejected
then in each case we have, = 6
= 4 2 . 2 1
if 3, 4, or 1 is not taken then 10 number for each case. Total = 30
Hence total = 30 + 6 + 6 = 42 ]
[22] Q.58/8 There are 12 guests at a dinner party. Supposing that the master and mistress of the house have fixed
seats opposite one another, and that there are two specified guests who must always, be placed next to one another ; the number of ways in which the company can be placed, is:
(A*) 20 . 10 ! (B) 22 . 10 ! (C) 44 . 10 ! (D) none
[Hint: 6 places on either sides G1G2 will have 5 places each on either side and can be seated in 2 ways 10 × 2! × 10! Ans ]
Q.69/8 Let Pn denotes the number of ways of selecting 3 people out of 'n' sitting in a row, if no two of them are consecutive and Qnis the corresponding figure when they are in a circle. If Pn Qn =6, then 'n' is equal to : (A) 8 (B) 9 (C*) 10 (D) 12 [Hint : Pn = n 2C3 ; Qn = nC3 [ n + n (n 4) ] or Qn = n n C1 C 4 2 3 . Pn Qn = 6 n = 10 ]
Q.7207/1(10/8) Define a 'good word' as a sequence of letters that consists only of the letters A, B and C and in which A never immidiately followed by B, B is never immediately followed by C, and C is never immediately followed by A. If the number of n-letter good words are 384, find the value of n. [Ans. n = 8 ] [Sol. There are 3 choices for the first letter and two choices for each subsequent letters.
Hence using fundamental principle number of good words = 3 · 2n–1 = 384
2n–1 = 128 = 27
n = 8 Ans. ]
Q.811/8Six married couple are sitting in a room. Find the number of ways in which 4 people can be selected so that (a) they do not form a couple (b) they form exactly one couple
(c) they form at least one couple (d) they form atmost one couple [Hint : 12C 4 = 6C 2 + 6C 1 · 5C 2 · 2 4 + 6C 4 · 2 4] [Ans. 240, 240, 255, 480]
Q.9 In a conference 10 speakers are present. If S1 wants to speak before S2 & S2 wants to speak after S3, then the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining seven speakers have no objection to speak at any number is
(A) 10C 3 (B) 10P8 (C) 10P3 (D*) 10 3 ! [Hint: S1 S2 S3 or S3 S1 S2 10C 3 . 2 . 7! ]
Q.102/9Let m denote the number of ways in which 4 different books are distributed among 10 persons, each receiving none or one only and let n denote the number of ways of distribution if the books are all alike. Then : (A) m = 4n (B) n = 4m (C*) m = 24n (D) none [Hint: m = 10C 4 . 4! and n = 10C 4 ]
Q.115/9The number of all possible selections of one or more questions from 10 given questions, each equestion having an alternative is :
(A) 310 (B) 210 1 (C*) 310 1 (D) 210 [Hint : 1st question can be selected in three ways and so on ]
Q.12 Number of 7 digit numbers the sum of whose digits is 61 is :
(A) 12 (B) 24 (C*) 28 (D) none
Q.139/9The number of ways of choosing a committee of 2 women & 3 men from 5 women & 6 men, if Mr. A refuses to serve on the committee if Mr. B is a member & Mr. B can only serve, if Miss C is the member of the committee, is
(A) 60 (B) 84 (C*) 124 (D) none
[Hint : ; ;
(i) Miss C is taken
(a) B included A excluded 4C 1 .
4C 2 = 24 (b) B excluded 4C
1 . 5C3 = 40 (ii) Miss C is not taken
B does not comes ; 4C 2 .
5C
3 = 60 Total = 124
Alt. Total [A, B, C present + A,B present & C absent + B present & A, C absent] Alternatively : Case 1 : Mr. 'B' is present 'A' is excluded & 'C' included
Hence number of ways = 4C
2 . 4C1 = 24 Case 2 : Mr. 'B' is absent no constraint
Hence number of ways = 5C 3 .
5C
2 = 100 Total = 124 ]
Q.1410/9 Six persons A, B, C, D, E and F are to be seated at a circular table. The number of ways this can be done if A must have either B or C on his right and B must have either C or D on his right is :
(A) 36 (B) 12 (C) 24 (D*) 18
[Hint : when A has B or C to his right we have AB or AC when B has C or D to his right we have BC or BD Thus we must have ABC or ABD or AC and BD
for D, E, F on a circle number of ways = 3! = 6 for C, E, F on a circle number of ways = 3! = 6
for , E, F the number of ways = 3! = 6 Total = 18 ]
Q.1511/9 There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the same colour, is : (A*) 6 (7! 4!) (B) 7 (6! 4!) (C) 8! 5! (D) none [Hint : 9 2 3 ! ! !
– number of ways when balls of the same colour are together = 9
2 3 !
! ! 3! 4! = 6(7! 4!) ]
Q.161/9Product of all the even divisors of N = 1000, is
(A) 32 · 102 (B) 64 · 214 (C*) 64 · 1018 (D) 128 · 106
[Sol. 1000 = 23 · 53 [11th, 28-11-2009, P-2] [Dpp, p&c]
Number of divisors = 16 which are (20 + 21 + 22 + 23)(50 + 51 + 52 + 53) Product of all the 16 divisors is
= (56)(56)(56)(56)(26)(26)(26)(26) = 224 · 524
Now odd divisors are 1, 5, 25, 125 and product of odd deivisors = 56
Product of all the even divisors = 224 · 524 = 218 · 518 · 26 = (1018)64 Ans.
Alternatively: Prodcut of all the even divisors = (24 · 56)(28 · 56) (212 · 56) (think!) = 224 · 518 = 64 · 1018 Ans. ]
[24] Q.17209/1(14/7) Find the number of 4 digit numbers starting with 1 and having exactly two identical digits.
[Ans. 432] [Sol. Case-I : When the two identical digits are both unity as shown.
any one place out of 3 block for unity can be taken in 3 ways and the remaining two blocks can be filled in 9 · 8 ways.
Total ways in this case = 3 · 9 · 8 = 216
Case-II : When the two identical digit are other than unity. ; ;
two x's can be taken in 9 ways and filled in three ways and y can be taken in 8 ways. Total ways in this case = 9 · 3 · 8 = 216
Total of both case = 432 Ans. ]
Q.18 Consider the word W = MISSISSIPPI
(a) If N denotes the number of different selections of 5 letters from the word W = MISSISSIPPI then N belongs to the set
(A) {15, 16, 17, 18, 19} (B) {20, 21, 22, 23, 24}
(C*) {25, 26, 27, 28, 29} (D) {30, 31, 32, 33, 34}
(b) Number of ways in which the letters of the word W can be arranged if atleast one vowel is separated from rest of the vowels
(A) ! 2 · ! 4 · ! 4 161 · ! 8 (B*) ! 2 · ! 4 · 4 161 · ! 8 (C) ! 2 · ! 4 161 · ! 8 (D) ! 4 165 · ! 2 · ! 4 ! 8
(c) If the number of arrangements of the letters of the word W if all the S's and P's are separated is (K) ! 4 · ! 4 ! 10 then K equals (A) 5 6 (B*) 1 (C) 3 4 (D) 2 3 [Sol. [13th, 28-12-2008, P-1] (a) S S S S ; I I I I ; P P ; M 4 alike + 1 different = 2 · 3 = 6
3 alike + 2 other alike = 2 · 2 = 4
3 alike + 2 different = 2 · 3 = 6
2 alike + 2 other alike + 1 different = 3 · 2 = 6
2 alike + 3 different = 3 = 3
25
(b) Atleast one vowel is separated from the rest
= total – when all vowels together = ! 2 · ! 4 · ! 4 ! 11 – ! 2 · ! 4 ! 8 = ! 2 · ! 4 · ! 4 ! 8 · 9 · 10 · 11 – ! 2 · ! 4 ! 8 = 1 24 9 · 10 · 11 ! 2 · ! 4 ! 8 = 1 4 15 · 11 ! 2 · ! 4 ! 8 = 4 4 165 ! 2 · ! 4 ! 8 = 4·4!·2! 161 · ! 8 Ans.
(c) Separate S first | I | I | I | I | P | P | M | (8 gaps) | I | I | I | I | PP | M | 4 7 4 8 C · ! 4 ! 6 C · ! 2 · ! 4 ! 7
; S's separated but P's may or may not be separated ;
simplifies to ! 4 · ! 4 ! 7 · ! 6 = ! 4 · ! 4 ! 10 K = 1 Ans. ]
Q.19 Let A = {a, b, c, d, e, f} and B = {1, 2, 3} are two sets.
Let m denotes the number of mappings which are into from A to B n denotes the number of mappings which are injective from B to A.
Find (m + n). [Ans. 309]
[Sol. m = into mappings from A to B
Case-I: If exactly one element of set B is not the image of any of the elements of set A then total number of into functions are
3C 1 × (2
6 – 2) = 3 × 62 = 186
Case-II: If exactly two elements of set B is not the image of any of the elements of set A then total number of into functions are 3C 2 × 1 = 3 and hence m = 186 + 3 = 189 [12th, 19-07-2009, P-1] Alternatively: Total mappings = 36 = 729 Surjective mappings 2 2 2 90 8 720 ! 3 ) ! 2 ( ! 3 ! 6 3 2 3 1 360 2 720 ! 3 ! 2 ! 3 ! 6 1 1 4 90 4 2 720 ! 4 ! 2 ! 3 ! 6
Total surjective mapping = 540 not surjective mappings = 729 – 540 = 189 = m n = number of injective mapping from B to A.
Injective mapping = 6C
3 × 3! = 20 × 6 = 120 = n (m + n) = 189 + 120 = 309 Ans. ]
[26]
Q.20 Let a function f is defined as f : {1, 2, 3, 4} {1, 2, 3, 4}. If f satisfy f
f(x)
= f (x), x {1, 2, 3, 4}then find the number of such function. [Ans. 41]
[Sol. f
f(x)
= f (x) ; f (x) = y f (y) = y [12th, 23-08-2009, P-2] Case-1: range contains exactly one elementit can be done in 4C
1 ways say 1
remaining 3 elements i.e. 2, 3, 4 can be mapped only in one ways total = 4C
1 = 4 Case-2: range contains two elements this can
this can be done in 4C
2 ways say 1, 2 f (1) = 1; f (2) = 2
remaining 2 elements i.e. 3 and 4 each can be mapped in 2 ways
Total = 4C
2 · 22 = 24
Case-3: range contains 3 elements which can be done in 4C
3 ways say 1, 2, 3 f (1) = 1; f (2) = 2 and f (3) = 3 now remaining 4 can be mapped only in 3 ways
Total = 4C
3 · 3 = 12 ways Case-4: range contains all 4 elements
f (1) = 4; f (2) = 2 ; f (3) = 3; f (4) = 4 only 1 way
MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
Choose the correct alternative (only one is correct):
Q.1 Number of ways in which four different toys and five indistinguishable marbles can be distributed between Amar, Akbar and Anthony , if each child receives atleast one toy and one marble, is
(A) 42 (B) 100 (C) 150 (D*) 216
[Sol. Toys in group 1 1 2 3! ! 2 ! 2 ! 1 ! 1 ! 4 = 36 Marbles OO O O = 4C 2 = 6 Total ways = 36 × 6 = 216 ]
Q.2 A 3 digit palindrome is a 3 digit number (not starting with zero) which reads the same backwards as forwards. For example 171. The sum of all even 3 digit palindromes, is
(A) 22380 (B) 25700 (C*) 22000 (D) 22400
[Sol. Number of even 3 digit palindromes = 4 · 10 = 40
U T H
[12th & 13th 03-03-2007] 4 ways to fill (hundredth and unit's place) and 10 ways to fill tenth's place
sum = ) place 100 ( th ) 8 6 4 2 ( 10 100 + ) place units of sum ( ) 8 6 4 2 ( 10 + ) place 10 ( th ) 9 ... 3 2 1 ( 10 · 4 = 20,000 + 200 + 1800 = 22000]
Q.33/10There are 100 different books in a shelf. Number of ways in which 3 books can be selected so that no two of which are neighbours is
(A) 100C
3 – 98 (B) 97C3 (C) 96C3 (D*) 98C3
[11th & 13th, 16-12-2007]
Q.4 A lift with 7 people stops at 10 floors. People varying from zero to seven go out at each floor. The number of ways in which the lift can get emptied, assuming each way only differs by the number of people leaving at each floor, is :
(A) 16C
6 (B) 17C7 (C*) 16C7 (D) none
[Hint : consider each floor to be a beggar. Now distribute 7 identical coins ( 7 people) in 10 beggars each receiving none, one or more 16C
7 ]
Q.5 You are given an unlimited supply of each of the digits 1, 2, 3 or 4. Using only these four digits, you construct n digit numbers. Such n digit numbers will be called L E G I T I M A T E if it contains the digit 1 either an even number times or not at all. Number of n digit legitimate numbers are
(A) 2n + 1 (B) 2n + 1 + 2 (C) 2n + 2 + 4 (D*) 2n – 1(2n + 1) [Sol. × × ×...× (n places) [12th test (09-10-2005)]
Total numbers = 3n + nC
2 · 3n – 2 + nC4 · 3n – 4 + ... (nC2 indicates selection of places)
= 2 ) 1 3 ( ) 1 3 ( n n = 2 1 [4n + 2n] = 22n – 1 + 2n – 1 = 2n – 1(2n + 1) ]
Q.65/10Two classrooms A and B having capacity of 25 and (n–25) seats respectively.An denotes the number of possible seating arrangements of room 'A', when 'n' students are to be seated in these rooms, starting from room 'A' which is to be filled up full to its capacity. If An – An–1 = 25! (49C
25) then 'n' equals (A*) 50 (B) 48 (C) 49 (D) 51 [Hint: Given An = nC 25 · 25! ; An – 1 = n–1C25 · 25! hence nC 25. 25! – n–1C 25 . 25! = 25! 49C 25
[28] or n–1C 25 + n–1C 24 – n–1C 25 = 49C 24 n – 1 = 49 n = 50 n–1C 24 = 49C 24 n = 50]
Q.79/10Number of positive integral solutions satisfying the equation (x1 + x2 + x3) (y1 + y2) = 77, is
(A) 150 (B) 270 (C*) 420 (D) 1024
[Sol. (x1 + x2 + x3) (y1 + y2) = 11 · 7 or 7 · 11 [29-01-2006, 12 & 13] In the first case (x1 + x2 + x3) = 11 and (y1 + y2) = 7, which have 10C
2 · 6C1 solutions (using beggar) In the second case (x1 + x2 + x3) = 7 and (y1 + y2) = 11, which have 6C
2 · 10C
1 solutions (using beggar) total number of solutions = 10C
2 · 6C1 + 6C2 · 10C1 = 270 + 150 = 420 Ans. ]
Q.810/10 Distinct 3 digit numbers are formed using only the digits 1, 2, 3 and 4 with each digit used at most once in each number thus formed. The sum of all possible numbers so formed is
(A*) 6660 (B) 3330 (C) 2220 (D) none
[Hint: all possible = 24
6(1 + 2 + 3 + 4)(1 + 10 + 102) = 6 · 10 · 111 = 6660
reject 1 or 2 or 3 or 4 ] [12 & 13th test (09-10-2005)]
Q.9 There are counters available in 3 different colours (atleast four of each colour). Counters are all alike except for the colour. If 'm' denotes the number of arrangements of four counters if no arrangement consists of counters of same colour and 'n' denotes the corresponding figure when every arrangement consists of counters of each colour, then :
(A) m = 2 n (B*) 6 m = 13 n (C) 3 m = 5 n (D) 5 m = 3 n [Hint : m = 34 3 = 78; n = 34
3 2
4 2
3
= 81 45 = 36 Hence m n = 78 36 = 13 6 6 m = 13 n B ]Q.1012/10 An ice cream parlour has ice creams in eight different varieties. Number of ways of choosing 3 ice creams taking atleast two ice creams of the same variety, is
(Assume that ice creams of the same variety to be identical & available in unlimited supply)
(A) 56 (B*) 64 (C) 100 (D) none [Hint: 10C 3 8C 3 = 120 56 = 64 ] [Note: 10C
3 = all diff. + 2 alike and 1 diff. + all alike]
Q.1113/10 There are 12 books on Algebra and Calculus in our library, the books of the same subject being different. If the number of selections each of which consists of 3 books on each topic is greatest then the number of books of Algebra and Calculus in the library are respectively:
(A) 3 and 9 (B) 4 and 8 (C) 5 and 7 (D*) 6 and 6
[Hint: = xC
3 · 12 – xC3 ]
Q.12 Three digit numbers in which the middle one is a perfect square are formed using the digits 1 to 9 . Their sum is :
(A*) 134055 (B) 270540 (C) 170055 (D) none of these [ Hint: Middle place 1, 4 & 9
Two terminal positions 1, 2, ... , 9
Hence total numbers = 9 . 9 . 3 = 243 (Terminal digits in 9 ways and middle one in 3 ways) For the middle place 1, 4 & 9 will come 81 times
sum = 81 10 (1 + 4 + 9) A
For units place each digit from 1 to 9 will appear 27 times sum = 27 (1 + 2 + ... + 9) B
For hundreath's place, similarly sum = 27 100 (1 + 2 + ... + 9) C A + B + C gives the required sum ]
Q.13 A guardian with 6 wards wishes everyone of them to study either Law or Medicine or Engineering. Number of ways in which he can make up his mind with regard to the education of his wards if every one of them be fit for any of the branches to study, and atleast one child is to be sent in each discipline is :
(A) 120 (B) 216 (C) 729 (D*) 540
[Hint: Divide 6 children into groups as 123, 411 ot 222
Now total = ! 3 ! 2 ! 1 ! 6 3! + ! 2 ! 1 ! 1 ! 4 ! 6 3! + ! 3 ! 2 ! 2 ! 2 ! 3 . ! 6 = 360 + 90 + 90 = 540 ]
Q.1415/8 There are (p + q) different books on different topics in Mathematics. (p q)
If L = The number of ways in which these books are distributed between two students X and Y such that X get p books and Y gets q books.
M = The number of ways in which these books are distributed between two students X and Y such that one of them gets p books and another gets q books.
N = The number of ways in which these books are divided into two groups of p books and q books then,
(A) L = M = N (B) L = 2M = 2N (C*) 2L = M = 2N (D) L = M = 2N
Q.15 Number of ways in which 5 A's and 6 B's can be arranged in a row which reads the same backwards and forwards, is
(A) 6 (B) 8 (C*) 10 (D) 12
[Sol. A A A A A | B B B B B B Middle digit must be A (think !)
M
so that even number of A's and B's are available
Take AABBB on one side of M (6th place) and then their image about M in a unique way
Number of ways = ! 3 · ! 2 ! 5 = 10 Ans.] [12th, 14-06-2009, P-2]
Q.16 A person writes letters to his 5 friends and addresses the corresponding envelopes. Number of ways in which the letters can be placed in the envelope, so that atleast two of them are in the wrong envelopes,is,
(A) 1 (B) 2 (C) 118 (D*) 119 [Hint: 5! = W 4 and R 1 + W 3 and R 2 + W 2 and R 3 + way one used none R 5 4 + all 5 wrong
hence at least 2 in wrong = 120 – 1 = 119 ]
Q.17 For a game in which two partners oppose two other partners, 8 men are available. If every possible pair must play with every other pair, the number of games played is
(A) 8C
2 . 6C2 (B) 8C2 . 6C2 . 2 (C*) 8C4 . 3 (D) none [Hint: For each game 4 persons are needed . Hence select 4 from 8 in 8C
4 way . Now from each selection 3 games can be had 8C
4 . 3 = 210 ]
Q.18 The number 916238457 is an example of nine digit number which contains each of the digit 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. Number of such numbers are
(A) 2268 (B*) 2520 (C) 2975 (D) 1560
[Sol. 9, 8 & 7 can be placed in 9 · 8 · 7 ways.
There are only five ways to place the 6 (any where except the right most remaining slot, think !) and the order of 1 - 5 is fixed.
