Permutations and Combinations

1. Determine the middle term in the expansion of

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To get the k-value for the middle term, divide the exponent by 2: 10= 5

2 .

Now apply the formula:

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2. In how many ways can 13 colleagues sit at two round conference tables if the first

table seats seven and the second table seats six?

To begin, select the seven colleagues that will sit at the first table. This can be done in ₁₃C =1716₇ ways.

Then arrange these seven people in a circle in

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in a circle in

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The answer is obtained by multiplying these results: 1716 × 720 × 120 = 148 262 400

3. Fred is having a party and invites 5 out of his 8 friends.

a) In how many ways can he choose which five people come to the party?

The order of friends attending the party is not important. Use a combination. ₈C = 56₅

b) In how many ways can he choose the five guests if Daniel and Stephanie must either

attend together or not at all?

If Daniel and Stephanie are together, that means out of 6 friends (two fewer remain in the group), Fred must select

three (since Daniel & Stephanie are already chosen) This can be done in ₆C = 20₃ ways.

If Daniel and Stephanie are not coming to the party, that means there are 6 people available to choose from,

and all 5 positions are still available. This can be done in ₆C =6₅ ways.

4. How many seven digit numbers are possible using 1, 2, 3, 4, 5, 6, and 7 if the middle

digit is even and repetitions are not allowed?

There are three even digits which may go in the middle: Now fill in the remaining spaces with the rest of the digits

5. There are three flights from Edmonton to Winnipeg, and five flights from Winnipeg to

Toronto. How many different ways can a passenger fly from Edmonton to Toronto through Winnipeg?

Multiply the first set of flights (3) by the second set of flights (5). 3 × 5 = 15

6. How many numbers between 100 and 1000 are odd? Repetitions are allowed. The last digit must be 1, 3, 5, 7, or 9 for the number to be odd.

Therefore, there are five possibilities for this position.

Now fill in the remaining positions. There are 9 possibilities for the first digit (it can’t be zero), and 10 possibilities for the second.

7. Determine the sixth term in the expansion of

9 2 3 x x ⎛ ₋ ⎞ ⎜ ⎟ ⎝ ⎠

Use the formula t_k+1= C a_{n k}

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8. In how many ways can letters from the word EDITOR be arranged if vowels and

consonants alternate positions?

First determine the number of arrangements with consonants first: Then determine the number of arrangements with vowels coming first:

Add the results to get 72 possible arrangements.

9. How many different arrangements of the word MATHEMATICS are there if the M’s

must stay together?

Put a circle around the two M’s to indicate they are one object:

Now evaluate the number of arrangements: 10! =907200 2!2!

(We must divide out the repeated A’s and T’s. We don’t need to divide out the two M’s since they have been combined into one object) 10. How many four digit numbers between 999 and 9999 are divisible by 5 and have

no repeated digits?

There are two separate cases that must be addressed. The first case involves a five being the last digit, and the second case involves a zero

being the last digit. Fill in the spaces as follows:

Now fill in the remaining spaces:

11. In the expansion of

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Use the formula t_k+1= C a_{n k}

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First predict what k has to be in order to get the required term:

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14 +1

t = C x y

If the empty box is filled with the number 11, you will get the required term.

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12. If there are six empty seats in a row at a movie, how many ways could four people fill

those seats?

Arrangement matters in this case, so use a permutation. ₆P = 360₄

13. Solve for n: _nP₂ =56 n 2 2 2 P = 56 n! ₌₅₆ (n- 2)! n(n-1)(n- 2)!_{= 56} (n- 2)! n(n-1)= 56 n - n= 56 n - n-56=0 (n-8)(n+7)=0 n=8

Reject -7 since you can’t have negative objects to select from. 14. The head of a committee needs to assign members to a sub-committee. The positions

of president, vice-president, and treasurer must be filled. In how many ways can Gary, John, Kevin, Ned, Stacy, and Veronica be assigned to these three positions?

Since titles are involved in the committee, use a permutation. There are 6 people available to fill three positions. ₆P =120₃

15. Mark has 25 songs and wishes to put 14 of them on a mix CD. a) In how many ways can the 14 songs be chosen?

Order does not matter if the songs are merely being chosen.

25C = 445740014

b) The songs “Corners” and “Predictions” sound very similar. If Mark uses no more than

one of these songs on his mix CD, how many was can he choose the 14 songs for his CD? Watch out for the wording “no more than one”.

That means he could choose neither song, or one song. The number of ways to select the 14 songs without picking either

“Corners” or “Predictions” is ₂₃C = 817170₁₄ . The number of ways to pick the 14 songs using

one of the songs is ₂C • C = 2288132_{1 23 13}

Add the results to get the total number of cases: 3105322 16. Using the letters from the word PENCILS,

a) Determine the number of 5 letter arrangements

7P = 2520 5

b) How many seven letter arrangements are possible if N must be the first letter and the

letters P & I must be together?

First group the letters as follows:

If N must go first, there is only one possibility for that position. Out of the 5 objects remaining (remember the bubble counts as only

one object), we can arrange them as shown.

However, don’t forget that we can still arrange what’s in the bubble in 2! = 2 ways.

17. There are 4 different paths from home to school, and 6 different paths from school to

the convenience store. How many ways can a student go from home to the convenience store, stopping at the school along the way?

Multiply the number of paths to the school (4) by the number of paths to the convenience store (6).

4 × 6 = 24

18. If there are three different cars and seven parking spaces, how many unique ways can

the cars be parked?

Use a permutation since order is important: ₇P = 210₃

19. Solve for n algebraically: (n+3)! 20(= n+1)!

2 2 (n+3)!= 20(n+1)! (n+3)(n+2)(n+1)!= 20(n+1)! (n+3)(n+2)= 20 n +5n+6= 20 n +5n-14=0 (n+7)(n- 2)=0 n= 2

Reject -7 since n must be a whole number

20. Given (3x + 2y)8, determine the middle term of the expansion. The k for the middle term is 8 ÷ 2 = 4

8-4 4 8 4 4 4 4 4 C (3x) (2y) =70(3x) (2y) =70(81x )(16y ) 4 4 =90720x y

21. A teacher wishes to have a picture taken with six team players. In how many ways

22. How many four digit positive numbers less than 4670 can be formed using the digits

1, 3, 4, 5, 8, 9 if repetition is not allowed.

We need to separate this question out into different cases since numbers in the 4000’s have extra restrictions.

Case 1: Numbers in the 4000’s: There is only one possibility for the first digit {4}. The next digit has three possibilities. {1, 3, 5}. There are 4 possibilities for the next digit since any remaining number can be

used, and 3 possibilities for the last digit.

Case 2: Numbers in the 1000’s and 3000’s: There are two possibilities for the first digit {1, 3}. Anything goes for the remaining

digits, so there are 5, then 4, then 3 possibilities. Add the results together: 36 + 120 = 156

23. In how many ways can the letters in the word QUILT be arranged if the vowels and

consonants alternate?

Since there are three consonants and two vowels, there is only one option. It must go CVCVC

24. Simplify: 400 300 400 100 C C

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25. Using all the letters from the word SELECTIONS, how many distinguishable

arrangements can be made?

There are 10 letters. There are 2 E’s repeated and two S’s repeated. The number of arrangements is 10! = 10! =

2!•2! 2!•2! 907200 26. Given (8x6 – 7y3)9 , determine the position of the term containing x36

First find the k value of the term containing x36

9 6 9- 3

36

C (8x ) (-7y ) By inspection, we'll get x when k = 3

This corresponds to the fourth term of the expansion since the k-value is always one less than the term position.

27. Cassie, Richard, and 7 of their friends attend a movie. In how many ways can they be

seated so that Cassie and Richard do not sit next to each other?

To find the number of ways Cassie and Richard can stand next to each other in line, treat Cassie & Richard as a single item.

The number of arrangements is 8!•2! = 80640.

(Don’t forget to use the 2! since Cassie & Richard can be arranged two different ways inside their bubble.)

The number of ways nine people can be arranged without restrictions is 9! = 362880

The number of ways Cassie and Richard do not stand next to each other in line is 362880 – 80640 = 282240

28. In how many ways can six people be seated at a round table?

Use the formula for a circle permutation

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29. The fifth term in the expansion of 4 3 n a a ⎛ − ⎜ ⎝ ⎠ ⎞

⎟ contains . Determine the value of n. a4 Use the formula t_k+1= C a_{n k}

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To get the fifth term, k = 4.

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⎝ ⎠ 4 n-4 4 4+1 n 4 3 t = C a -a

By inspection, we can see that n = 6 yields the correct exponent for a.

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30. Solve for n in the equation ( 2)! 8

( 1)! n n + ₌ + (n+ 2)!_{= 8} (n+1)! (n+ 2)(n+1)!_{= 8} (n+1)! n+ 2= 8 n=6

31. How many terms are in the expansion of

7 3 1 x x ⎛ ₊ ⎞ ⎜ ⎟ ⎝ ⎠

The number of terms in an expansion is one more than the value of n. Therefore, there are 8 terms.

32. In how many ways can six students sit in a row if two students, Jerry and Jennifer,

must sit together?

33. Find and simplify the fourth term in the expansion of

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To get the fourth term, k = 3.

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34. How many four letter arrangements can be made from the letters DDEFGH This question is tricky since we have to look at three cases. The case where no D is chosen: Here we wish to arrange the letters

EFGH, which can be done in 4! = 24 ways.

The case where one D is chosen: If a D is selected, then we need to choose three additional letters from EFGH. This can be done in ₄C₃ ways. Now that we have selected all four letters, arrange them in 4! ways. The number of cases where one D is selected is

4C ×4!=963

The case where two D’s are chosen: If two D’s are selected, then we still need to select two letters from EFGH. This can be done

in ₄C₂ ways. Now that we have chosen all four letters, arrange them in 4! ways. Don’t forget that since we have two D’s,

we need to divide out repetitions!

The number of cases where two D’s are selected is ₄C ×₂ 4!=72

2! .

Add up all the separate cases: 24 + 96 + 72 = 192

*If you thought you can use 6 4P

2! , this is incorrect because it only works if repeated letters are present in

35. The term −1080a b2 3 occurs in the expansion of

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First predict what k has to be in order to get the required term: To get back b3_{, there is only one option for k. It has to be 3.}

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t = C a b

To get back a2_{, the value of n must be 5}

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36. There are 15 musicians to be separated into three groups of five. The players for

Group 1 are chosen first, then Group 2 and finally the Group 3. How many ways can this be done?

The players for Group 1 can be chosen in ₁₅C = 3003₅ ways. The players for Group 2 can be chosen in ₁₀C = 252₅ ways,

since only 10 people remain to be selected.

The players from Group 3 can be chosen in ₅C = 1₅ way. The total number of ways is 3003 × 252 × 1 = 756756 ways. 37. A grad committee of 5 students is to be formed from a class of 21 students.

Danielle, Francis, and Hillary decide they do not want to be on the committee. How many committees can be formed? (Express answer as factorials)

If these three people are not going to be on the committee, they can be removed from the selection pool.

As a result, there are 18 students still in the selection pool, and all 5 positions must be filled.

This can be done in

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38. The expression

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2 a−

2

Use the formula t_k+1= C a_{n k}

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Set it up with the known information: t_k+1= C a_{11 k}

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The term containing a2 _{is the tenth term.}

39. If _nP_r =6720 and _nC_r =56, then the value of r is Write _{n r}P =6720 as

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Write _nC = 56_r as

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Now divide the expressions to simplify

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n!

n- r ! n! n- r !r! 6720

= × = r! On the right side, = 120 n! _{n- r ! n! 56}

n- r !r!

r! = 120 when r = 5. The answer is 5.