Permutations and Combinations

PERMUTATIONS AND

COMBINATIONS

by :

DR. T.K. JAIN

AFTERSCHO☺OL

centre for social entrepreneurship sivakamu veterinary hospital road

bikaner 334001 rajasthan, india

FOR – PGPSE / CSE PARTICIPANTS

My words...

My purpose here is to give a few questions on fundamentals of permutations and combinations. I welcome your suggestions. I also request you to help me in spreading social entrepreneurship across the globe – for

which I need support of you people – not of any VIP. With your help, I can spread the ideas – for which we

What is permutation &

combination?

Permutation = it denotes order / Sequence but combination = it only denotes that some

objects are together

example : ABC can have only one combination taking all of them together. But permutations are many : - ABC,ACB,BCA,BAC,CBA,CAB

What is formula of permutation ?

Npr = n! / (n-r)! p=permutation

n= total number of objects

r=how many objects you are taking at a time

! = multiply with reducing numbers till it reaches 1 example : 5p5 = 5! / (5-5)!

There are 5 books on maths, 3 on

entrepreneurship and 2 on economics. In how many ways can we arrange them. We have to

keep books of a subject together.

There are 3 subjects so we can arrange them in 3! methods = 6

there are 5 books on maths, we have 5! methods = 120 there are 3 books on entrepreneurship, so we have 3!

methods

there are 2 books on economics, so we have 2! methods. Total we have : 6*120*6*2 = 8640 answer

A court has given 6 to 3 decision

in favaour of an issues. In how

many ways can it reverse the

decision?

There are 9 judges, 5 will make a majority there may be 5,6,7,8,9 judges against the issue

we have 9c5 or 9c6 or 9c7 or 9c8 or 9c9 methods : 126 + 84+36+9+1 =256 answer

Three persons go into a room

which has 8 seats, in how many

ways can they occupy the seats ?

First person has 8 options, 2nd _{person has 7}

options and 3rd _{person has 6 options.}

There are 5 routes between Bikaner and Delhi. You may choose any route. In how

many ways can you go and return from Delhi?

How many permutations are

possible from ACCOUNTANT ?

There are 10 digits, there are 2A, 2C, 2T, 2N, so we have :

10! / (2!*2!*2!*2!) answer

How many different 4 digit letters

can you make out of A,B,C,D,E?

N = 5 (A,B,C,D,E) R = 4

formula = Npr = n! / (n-r)! =5!/(5-4)!

How many different 4 digit numbers can you make out of 1,2,3,4,0?

N = 5 (1,2,3,4,0) R = 4

but 0 cannot come in the first digit

for first digit we have 4 options (1,2,3,4), for next digits, we can use 0. thus we have 4*4*3*2*1 = 96 options OR

formula = Npr = n! / (n-r)!

=5!/(5-4)! but this contains all those numbers which start with 0. so let us keep 0 as fixed for 1st _{digit and solve it.}

contd...

If it is not 0, permutation will be : formula = Npr = n! / (n-r)!

=5!/(5-4)! = 120

Zero fixed for 1st _{potion, we have these options :} Npr = n! / (n-r)!

n=4,r=3 4!/(4-3)! = 24

How many different 4 digit numbers can you make out of 1,2,3,4,0 which are divisible by 2?

Start with 96 of the last question

now pick up all those which are ending with 1 : 3*3*2*1 = 18

similarly those which are ending with 3 3*3*2*1 = 18

In how many ways can Raj invite

any 3 of his 7 friends?

This is a question of combination. Here order

(sequence) is not important, his friends can come in any order. Thus this is a case of combination.

Formula : N! / ((n-r)!*r!)

you can calculate combination by dividing permutation by r!

How many different words can

you frame from FUTURE ?

Here we have two U total we have 6 digits.

Formula : N ! / L!

N= total number of digits

L = those digits which are repeated. Answer = 6! / 2!

How many different words can

you frame from DALDA ?

Here we have two D & A total we have 5 digits.

Formula : N ! / L!

N= total number of digits

L = those digits which are repeated. Answer = 5! / (2!*2!)

In how many ways can 8 person

sit around a round table ?

For questions relating to round table , we have to use the following formula :

(n-1)!

How many 4 digit numbers can be

formed out of 1,2,3,5,7,8,9 if no

digit is repeated.

Total number ofdigits = 7 formula = Npr

n =7 r 4 7p4 = 7! / 3! =7*6*5*4 = 840

How many numbers greater than

2000 can be formed from

1,2,3,4,5. No repeatition is

allowed.

5 digit numbers = 5! = 120

4 digit numbers,: we cant take 1 in the

beginning. We have 4 options for 1st _{digit 4 for}

2nd digit 3 for 3rd digit ... 4*4*3*2*1 = 96

There are 6 books on english, 3 on maths, 2 on GK. In how many ways can they be placed in

shelf, if books of 1 subject are together?

We have 3 subjects so 3!

books of same subjects can be interchanged. So answer : 3!*6!*3!*2!

How many words can we make

out of DRAUGHT, the vowels are

never separated?

Number of vowels = 2 other digits = 5

we will treat vowels as 1 word

so we have 6!. Vowels can be interchanged so 2!

In how many ways can 8 pearls be

used to form a necklace ?

In questions of necklace, we use the following formula : ½ (N-1)!

Here we can take reverse order of left to right or right to left, so divide by ½

=1/2 (8-1)! =2520

In how many number of ways can

7 boys form a ring ?

(7-1) !

50 different jewels can be set to

form necklace in how many

ways ?

½ ( n -1) ! = ½ (50 -1)!

How many number of different

digits can be formed from

0,2,3,4,8,9 between 10 to 1000?

Let us assume that repeatition is not allowed Let us make 2 digit numbers :

for first digit we have 5 option, for 2nd _{digit also we have 5 options} (including 0) = 25

for 3 digit numbers : 5*5*4 = 100 total 125

if repeatition is allowed :

for 2 digit : 5 * 6 = 30 for 3 digit : 5*6*6 = 180 total = 210 answer

What is the number of permutations of 10 different things taking 4 at a time in which one thing never

comes ?

= 9 p 4

= (9*8*7*6) =3024

There are 5 speakers (A,B,C,D,E) , in how many ways can we arrange their speach that A

always speaks before B

For A and then B without gap : Let us take A and B as one.

4! = 24

for A and then B let us keep B at 3rd _{place and A at 1}st place =3!

there are total 6 such possibilities so we have 6*6 = 36 total possibilities = 60 answer

5 persons are sitting in a round table in such a way that the tallest person always sits next to

the smallest person?

Keep tallest and smallest person as 1. we have (4-1)! = 6

the tallest and the smallest person can be interchanged = 2

How many words can be formed

from MOBILE so that consonent

always occupies odd place ?

There are 3 odd and 3 even places. We have 3! *3!

In how many ways can we

arrange 6 + and 4 – signs so that

no two – signs are together?

+ + + + + +

there are 5 places between 2 +. one on extreme left and one on extreme right.

We have 7 positions for – sign 7c4

There are 10 buses between Bikaner and Jaipur. In how many ways can Gajendra go to Jaipur and come

back without using the same bus in return journey?

There are 10 options while going

there are 9 options while returning (one bus used earlier will not be used)

In how many ways can yamini distribute 8 sweets to 8 persons provided the largest sweet

is served to Jigyasha?

1 sweet is fixed so we have 7! = 5040 answer

Yamini & Jigyasha go to a train and they find 6 vacant seats. In how many ways can they sit?

Yamini has 6 options but Jigyasha has only 5 options left

How many words can you make

from DOGMATIC?

8!

Gajendra has 12 friends out of whom 8 are relatives. In how many ways can he invite 7

in such a way that 5 are relatives?

8c5 * 4c2 =56*6

There are 8 points on a plane. No 3 points

are on a straight line. How many

traiangles can be made out of these ?

8c3

In how many ways can you form a

committee of 3 persons out of 12

persons ?

12c3

How many different factors are

possible from 75600 ?

The factors are : 2^4* 3^3*5^2 *7

formula = (number of factors +1) (number of factors +1) .... - 1

A box contains 7 red 5 white and 4 blue balls. How many selections can be made that we pick

up 3 balls and all are red?

It is a question of combination. Total possibilities = 7c3

7c3 = 7*6*5 / 3*2*1 = 35 thus there are 35 chances of getting

A box contains 7 red 5 white and 4 blue balls. What is the probability that in our selections

we pick up 3 balls and all are red? Total possibilities for red = 7c3

7c3 = 7*6*5 / 3*2*1 = 35

total possibility of 3 balls : 16c3 =(16*15*14/3*2*1) =560

probability

What is the probability of getting

3 heads when I toss a coin 5

times?

This is a case of binomial probability (where there are only 2 outcomes possible, we can use

this theory)

Here we can use this formula : Ncr (p)^r * (q)^(n-r)

=n =5, p = ½ q = (1-p) = ½ , r = 3 5c3 (1/2)^3*(1/2)^2

In how many ways can Gajendra invite some or all of his 5 friends in party hosted by him?

(at least 1)

Frmula of combination of 1 to all = 2^n – 1

= 2^5 - 1 = 32-1 =31 answer

How many words can be formed by using all the letters of the word DRAUGHT so that a. vowels always come together & b. vowels are never together?

A There are 2 vowels. We treat them as 1. solution : 6!*2! = 1440 answer

b. total possibilities = 7! = 5040

number of cases when vowels are not together = 5040-1440 = 3600 answer

In how many ways can a cricket

eleven be chosen out of a batch

of 15 players.

15c11 =15! / ((15-11)!*11!) =15!/(4!*11!) =(15*14*13*12)/(4*3*2*1) 1365 answer

In how many a committee of 5 members can be selected from 6 men 5 ladies consisting of 3 men and

2 ladies

6c3 *5c2

=[(6*5*4)/(3*2*1)] [(5*4)/(2*1)]

=20*10

How many 4-letter word with or without

meaning can be formed out of the letters of the word 'LOGARITHMS' if repetition of letters is

not allowed

10p4

=(10*9*8*7) =5040 answer

how many ways can the letter of

word 'LEADER' be arranged

We have two e, so divide 6p6 by 2 6!/2!

=720 / 2 =360 answer

How many arrangements can be made out of the letters of the word 'MATHEMATICS' be arranged

so that the vowels always come together

Let us treat all 4 vowels as 1 total digits are 11

we we take 11 – 4+1 = 8 digits

vowels can be arranged among themselves = 4!/2!

In how many different ways can the letter of the word 'DETAIL' be

arranged in such a way that the vowels occupy only the odd positions

We have 3 odd and 3 even positions =3! *3!

How many 3 digit numbers can be formed from the digits 2,3,5,6,7 and 9 which are divisible by 5 and

none of the digits is repeated?

Last digit must be 5

now we have 5 options for 1st _{and 4 options for}

2nd digit

In how many ways can 21 books on English and 19 books on Hindi be placed in a row on a self so that

two books on Hindi may not be together?

We have 22 places for Hindi books. 22p19 *21!

Out of 7 constants and 4 vowels

how many words of 3 consonants

and 2 vowels can be formed?

Selection of 5 digits =7c3 *4c2

=35*6 = 210

5 digits can be arranged in 5! ways =120

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