Permutations and Combinations
Combinatorics
Copyright
Standards
© 2006,
Test - ANSWERS
Barry Mabillard.
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1. Determine the middle term in the expansion of ( a − 2b )
10
10
=5 .
2
Now apply the
formula:
n-k
k
tk+1 =n Ck ( a) ( b )
To get the k-value for the middle term, divide the exponent by 2:
t5+1 =10 C5 ( a)
10-5
(-2b )
5
t6 = 252a5 (-32b5 )
t6 = -8064a5 b 5
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 13 C7 = 1716 ways.
Then arrange these seven people in a circle in (7 - 1) ! = 6! = 720 ways.
Finally, take the six remaining people, and arrange them
in a circle in ( 6 - 1) ! = 5! = 120 ways.
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. 8 C5 = 56
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 6 C 3 = 20 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 6 C5 = 6 ways.
Add the cases to get 26 ways.
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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.
3⎞
⎛
7. Determine the sixth term in the expansion of ⎜ x 2 − ⎟
x⎠
⎝
Use the formula tk+1 =n Ck ( a)
9
( b) . For the sixth term, k = 5
n-k
k
tk+1 =n Ck ( a) ( b )
n-k
t5+1 =9 C5 ( x
k
)
2 9-5
⎛ 3⎞
⎜- ⎟
⎝ x⎠
⎛ 3⎞
t6 = 126 ( x ) ⎜ - ⎟
⎝ x⎠
⎛ 243 ⎞
t6 = 126x 8 ⎜ - 5 ⎟
⎝ x ⎠
t6 = -30618x 3
2 4
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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:
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)
Now evaluate the number of arrangements:
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:
Add the results to get the total: 952
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11. In the expansion of ( x + y ) , what is the numerical coefficient of the term
14
containing x3 y11
Use the formula tk+1 =n Ck ( a)
n-k
( b)
k
.
First predict what k has to be in order to get the required term:
t
+1
=14 C
(x)
14-
(y )
If the empty box is filled with the number 11,
you will get the required term.
14-11
11
t11+1 =14 C11 ( x )
(y )
t12 = 364x 3 y 11
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.
6
P4 = 360
13. Solve for n: n P2 = 56
n
P2 = 56
n!
= 56
(n- 2)!
n(n- 1)(n- 2)!
= 56
(n- 2)!
n(n- 1)= 56
n2 - n= 56
n2 - 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. 6 P3 =120
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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.
25 C14 = 4457400
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 23 C14 = 817170 .
The number of ways to pick the 14 songs using
one of the songs is 2 C1 •23 C13 = 2288132
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
7
P5 = 2520
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.
The answer is 120 × 2 = 240 ways.
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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:
7
P3 = 210
19. Solve for n algebraically: (n + 3)! = 20(n + 1) !
(n+3)! = 20(n+1)!
(n+3)(n+ 2)(n+1)! = 20(n+1)!
(n+3)(n+ 2)= 20
n2 +5n+6 = 20
n2 +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
C4 (3x)8-4 (2y)4
=70(3x)4 (2y)4
=70(81x 4 )(16y 4 )
=90720x 4 y 4
21. A teacher wishes to have a picture taken with six team players. In how many ways
can they line up for the picture if the teacher must stand in the middle?
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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:
C300
400 C100
400
C 300
400 C100
400
=
400!
( 400 - 300 ) !300!
=
400!
( 400 - 100 ) !100!
( 400 - 100 ) !100!
400!
×
400!
( 400 - 300 ) !300!
400!
300! 100!
×
100! 300!
400!
=1
=
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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.
10!
10!
=
= 907200
The number of arrangements is
2!• 2! 2!• 2!
26. Given (8x6 – 7y3)9 , determine the position of the term containing x36
First find the k value of the term containing x36
9
C (8x 6 )9- (-7y 3 )
By inspection, we'll get x 36
when k = 3
This corresponds to the fourth term of the expansion since the kvalue 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 ( n - 1) !
The answer is ( 6 - 1) ! = 5! = 120
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n
3⎞
⎛
29. The fifth term in the expansion of ⎜ a 4 − ⎟ contains a4 . Determine the value of n.
a⎠
⎝
Use the formula tk+1 =n Ck ( a)
n-k
( b)
k
To get the fifth term, k = 4. t4+1 =n C4 ( a4 )
.
n-4
⎛ 3⎞
⎜- ⎟
⎝ a⎠
4
By inspection, we can see that n = 6 yields the correct exponent for a.
4
4 6-4 ⎛ 3 ⎞
t5 =6 C 4 ( a ) ⎜ - ⎟
⎝ a⎠
2⎛ 3⎞
t5 = 15 ( a4 ) ⎜ - ⎟
⎝ a⎠
⎛ 81 ⎞
t5 = 15a8 ⎜ - 4 ⎟
⎝ a ⎠
t5 = -1215a4
30. Solve for n in the equation
4
Extra steps to
show the correct
exponent is
obtained.
(n + 2)!
=8
(n + 1)!
(n+ 2)!
=8
(n+1)!
(n+ 2)(n+1)!
=8
(n+1)!
n+ 2 = 8
n=6
1⎞
⎛
31. How many terms are in the expansion of ⎜ x3 + ⎟
x⎠
⎝
7
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?
We can arrange the row in 5!• 2! = 240 ways.
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33. Find and simplify the fourth term in the expansion of ( 2a − 3b )
Use the formula tk+1 =n Ck ( a)
n-k
( b)
k
6
.
To get the fourth term, k = 3.
6-3
3
t3+1 =6 C3 ( 2a) (-3b )
t4 = 20 ( 2a ) (-3b )
3
3
t4 = 20 ( 8a3 )(-27b3 )
t 4 = -4320a3 b 3
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 4 C3 ways. Now that we have selected all four letters, arrange
them in 4! ways. The number of cases where one D is selected is
4 C 3 ×4! =96
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 4 C2 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!
4!
The number of cases where two D’s are selected is 4 C 2 × =72 .
2!
Add up all the separate cases: 24 + 96 + 72 = 192
*If you thought you can use
6 P4
, this is incorrect because it only works if repeated letters are present in
2!
all arrangements. Obviously an arrangement like DFGH would not need the division!
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35. The term −1080a 2b3 occurs in the expansion of ( 2a − 3b ) The value of n is:
n
Use the formula tk+1 =n Ck ( a)
n-k
( b)
k
.
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.
n-3
3
t3+1 =n C3 ( a) ( b )
To get back a2, the value of n must be 5
5-3
3
t3+1 =5 C3 ( 2a) (-3b )
t4 =5 C3 ( 2a)
5-3
(-3b )
3
t4 = 10 ( 4a2 )(-27b3 )
t4 = -1080a2 b 3
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 15 C5 = 3003 ways.
The players for Group 2 can be chosen in 10 C5 = 252 ways,
since only 10 people remain to be selected.
The players from Group 3 can be chosen in 5 C5 = 1 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.
18!
18!
ways.
This can be done in 18 C 5 =
=
( 18 - 5 ) !5! 13!5!
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38. The expression ( a − 2 ) is expanded using descending powers of a. Determine the
11
position of the term containing a2
Use the formula tk+1 =n Ck ( a)
n-k
( b)
k
.
Set it up with the known information: tk+1 =11 Ck ( a)
11-k
(-2 )
k
To get back a2, there is only one option for k. It has to be 9.
11-9
9
t9+1 =11 C9 ( a )
(-2 )
t10 = 55a2 (-512 )
t10 = -28160a2
The term containing a2 is the tenth term.
39. If n Pr = 6720 and n Cr = 56 , then the value of r is
Write n Pr = 6720 as
Write n Cr = 56 as
n!
= 6720
( n- r ) !
n!
= 56
( n- r ) !r!
Now divide the expressions to simplify
n!
( n- r ) ! = n! × ( n- r ) !r! = r!
n!
n!
( n- r ) !
( n- r ) !r!
On the right side,
6720
= 120
56
r! = 120 when r = 5.
The answer is 5.
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