Download Pre-Calculus 12: Permutations and Combinations – Extra Review

Pre-Calculus 12: Permutations and Combinations – Extra Review
10
1.

1
th
 2x   ? What is the 10 term?
y


How many terms are in the expansion
2. A bowl contains an apple, a pear, a plum, and a banana. How many different pairs of fruit can be selected
from the bowl?
3. A combination lock that has 60 numbers on the dial works by turning it first to the right, then to the left,
and then to the right, with 3 selected numbers needed to open the lock. Is this process is an example of a
permutation, combination, both combination and permutation, or neither combination nor permutation?
4. Solve for n if n C3  120
5. There are 15 multiple-choice questions on an exam with 4 possible answers for each question.
a) How many different ways are there to complete the test?
b) How many different ways are there to finish the test if 5 answers are A, 3 are B, 4 are C, and 3 are D?
6. Find the constant term in (3 x 
2 12
) .
x3
7. A breakfast special consists of choosing one item from each category in the following menu.
Juice: apple, orange, grapefruit
Eggs: scrambled, fried, poached
Toast: white, brown
Beverage: coffee, tea, milk
a) How many different breakfast specials are possible?
b) JC doesn’t drink coffee and has to have scrambled eggs. How many breakfast combos does she have?
8. North American area codes are three digit numbers. Before 1995, area codes had the following
restrictions: the first digit could not be 0 or 8, the second digit was either 0 or 1, and the third digit was
any number from 1 through 9 inclusive. Under these rules, how many different area codes were possible?
9. Find the term containing x
44
in
( x 4  2 x 2 )15
10. Katie wants to colour a rainbow. She knows the 7 colors that make up a rainbow, but can’t remember the
right order. How many different ways could the colours be arranged assuming each colour is used only once?
11. Find the 4th term in the expansions of
(2 x 2  3 x)9 .
12. A chain of stores is offering a special price on a complete set of components (receiver, cassette deck,
speakers, and CD player) A purchaser is offered a choice of manufacturer for each component:
Receiver:
Kenwood
Pioneer
Sansui
Sony
Sherwood
Cassette Deck:
BSR
Dual
Sony
Technics
Speakers:
AR
KLH
JBL
CD Player:
Advent
Sony
Teac
Technics
a) In how many ways can one component of each be selected?
b) In how many ways can components be selected if both the Receiver & the Cassette Deck are to be Sony?
c) In how many ways can components be selected if none are to be Sony?
d) In how many ways can a selection be made of at least one Sony component is to be included?
13. Simplify the following expression without using the factorial symbol
1

14. The 10 term of the expansion of  x  
2

th
n
is 
(n  2)!( n  1)!
( n !) 2
1001 5
x . Determine n.
256
15. Linda and Sam play a tennis match. The first person to win 2 games wins the match. In how many different
ways can a winner be determined?
16. How many 6 digit numbers greater than 800 000 can be made from the digits 1,1,5,5,5,8?
17. Find the 6th term in the expansions of (
3
 2 x 4 )7 .
2
x
18. In how many ways can four colas, three iced teas, and three orange juices be distributed among ten
graduates if each graduate is to receive one beverage?
10
 1

 x3  is a constant?
2
 2x

19. Which term in the expansion of 
A
20. Moving only to the right or down, how many different routes exist to get
from point A to point B?
B
21. Sears wants to build 8 new stores in western Canada. They have the following information. If Sears wants
to study all possibilities for the locations of the 8 new stores, how many different possibilities would the
company have to consider?
Province
Number of stores to be built Number of possible locations
BC
2
6
Alberta
3
5
Saskatchewan
1
4
Manitoba
2
5
22. Numbers are formed on a calculator using seven lines which are either lit or not lit.
The diagram beside shows the number 8 formed using all 7 lines lit. How many
different symbols can be created by lighting one or more of these 7 lines?
( Count all the symbols, not just the ones that represent numbers.)
23. There are five boys and six girls on a grad committee.
a) In how many ways can a sub-committee of two boys and two girls be selected from the committee?
b) In how many how many ways can a sub-committee of four people be selected if there must be at least
one girl on the sub-committee?
c) In how many ways can a sub-committee of three boys and 2 girls be selected if Jack and Jill must be on
the committee?
d) In how many ways can a sub-committee of three boys and 2 girls be selected if either Jack or Jill, but
not both, must be on the committee.
e) In how many ways could a sub-committee of three boys and three girls be seated around a round table?
f) In how many ways can a sub-committee of three boys and 3 girls be selected and seated on a bench so
that the boys and girls alternate?
24. A checkerboard is an 8 x 8 game board. Game pieces can travel only diagonally on
the dark squares, one diagonal square at a time, and only in a downward direction.
If a checker is placed as shown, how many possible paths are there for the checker
to reach the opposite side of the game board?
25. Find the constant term in (6 x 
1 12
) .
4 x3
26. In how many ways can the digits 1,2,3,4,5,6,7 be arranged altogether, so that the even digits are always in
ascending order?
27. From five different calculus books, 4 different geometry books and seven different algebra books,
a) In how many ways can 3 calculus, 2 geometry and 4 algebra books be selected and arranged on a shelf?
b) In how many ways can 4 calculus, 3 geometry and 5 algebra books be selected and arranged on a shelf if
the calculus books must be kept together?
28. Solve for n.
10
Cn  120
29. Find the value of n if the numerical coefficient of x in the expansion of
3
( x  1)n is four times the
numerical coefficient of x .
2
30. How many ways can a committee of five be chosen from 10 girls and eight boys so that the girls have a
majority on the committee?
31. Solve for n.
5( n  2 ) P3  56n P2
32. A committee of five is to be selected from nine persons. In how many ways can this selection be made if:
a) Jack must be on the committee
b) Jack must not be on the committee
33. How many “words” of five different letters can be made from the English alphabet if each word: (NOTE:
Treat Y as a consonant)
a) consists of two vowels and three consonants
b) contains at most three consonants
c) contains at least one consonant and one vowel
d) begins with a consonant and ends in a vowel
34. Using the letters in the WEDNESDAY,
a) Find the number of arrangements that begin with the letter D
b) Find the number of arrangements that begin with exactly one D
c) Find the number of arrangements that have the letters DAY grouped together.
d) Find the number of arrangements that have the vowels grouped together.
e) Find the number of arrangements that have the consonants grouped together.
f) Find the number of 4 letter arrangements of the word WEDNESDAY that are possible.
35. Find the term containing x
44
in
( x 4  2 x 2 )15
36. Four couples attend a movie together and sit in a row with eight seats. In how many ways can they be
seated if:
a) each girl sits beside her date
b) the boys and girls alternate positions
37. Solve
P4  n  1C5 for n.
n
38. A True / False test has 15 questions on it.
a) How many different answer keys are possible?
b) How many different answer keys have 3 answers that are True and 12 answers that are False?
39. Deck of Cards
a) How many different five-card hands can be dealt from an ordinary deck of 52 cards?
b) How many of these hands contain five cards of the same suit?
c) How many of these hands contain cards with only even numbers on them?
d) How many of these hands contain cards with only numbers 2 through 10 or are red?
40.
How many five-card hands are possible consisting of:
a) exactly one pair and 3 different cards?
b) 4 diamonds and a heart?
c) exactly one pair, one queen, one king?
d) exactly two pairs?
e) exactly two pairs and a ten?
f) at least one club?
g) 3 of a kind and a pair?
h) a red 10 and a black 10 and three of a kind?
i) two hearts and 3 other cards all from another suit.
Solutions
1.
11 terms
20.
t10 10 C9 xy9
 1
 10(2 x)  
 y
 20 x

y9
4
3.
permutation
n!
 120
(n  3)!3!
4.
6.
t 4 12 C3 x y
4
1
2
4
C2 5 C3 4 C15 C2  6000
7
b) all - none
( x 4 )15  x 60 ,
x   x 
4 13
2 2
14 C2 5 C2  60
e)
t 4  9 C3 x y
1
2 1
58
,
so the x’s go down by 2.
5
4
5
9
5
x 44 term.
25.
x11 
x12 ,
14
28
5 x 4 x 3 x 4 = 240
1 x 1 x 3 x 4 = 12
4 x 3 x 3 x 3 = 108
all – none =
240 – 108 = 132
constant term.
t 4 12 C3 x 9 y 3
27.
a)
28.
1
64 x 9
38.
so the x’s go down by 2. So
t k 1  n C k x n  k y k
15.
Linda: WW,
WLW, LWW
3ways and the same pattern for Sam so there are
6 ways in total.
16. You must start with 8, 1 x 5 numbers (with
C3  4 C2 7 C4  9! 762048000
5
C4 4 C3 7 C5  420
n  7,3
Cn 3 4n Cn  2
n!
4n!

3!(n  3)! 2!(n  2)!
n
5!
 10
3!2!
3G2B & 4G1B & 5G
C3 8 C2 10 C4 8 C1 10 C5  5292
5(n  2)!
56n!

(n  1)! (n  2)!
5(n  2)( n  1)  56(n  1)
t6  7 C5 x 2 y 5
5n 2  41n  66  0
n  6 ( n  2.2 is not a solution)
a) Jack and 4 others
18 C4  70
 3
 21 2  (2 x 4 )5
x 
32.
9
 21  4  32 x 20
x
33. AEIOU – 5 vowels 21 consonants
a)
C  C  5! 1596000
2
 6048x16
b) not Jack & 5 others
5
b)
10!
 4200
4!3!3!
10
 1 31 1
 2  ( x )  15 ,
x
x 
8
1
 1 3 2
 2  ( x )  10 ,
x
x 
So the 5th term will be the constant term.
c)
2 21
8
C5  56
3
5 C5 21 C0  5! 120
5 C4 21 C1  5! 12600
5 C3 21 C2  5! 252000
5
9
C2 21C3  5! 1596000
total = 1 860 720
all – just vowels or consonants
P  5!21P5  5451600
26 5
d)
a)
consonant 24others vowel
21 x
P x 5 = 1 275 120
24 3
215  32768
15!
 455
3!12!
b)
C5  2598960
C
4 1 13 C5  5148
c)
20
C5  15504
d)
44
C5  1086008
a)
40.
a)
13
b)
13
52
C1 4 C2 12 C3 4 C1 4 C1 4 C1  1098240
d)
C4 13 C1  9295
C
11 1  4 C2  4 C1  4 C1 10 C1  4 C1  42240
13 C2  4 C2  4 C2 11 C1  4 C1  123552
e)
12
f)
10
31.
39.
c)
1
2

 n  14
6 (n  2)
30.
x 44 term.
8
n!
(n  1)!

(n  4)! (n  4)!5!
5! n  1
120  n  1
119  n
b)
5
10!
 120
(10  n)!n!
10!
 (10  n)!n!
120
,
36. a) 4! x 2!2!2!2! = 384
b) 2 x 4!4! = 1152
37.
 34642080
7!
 840
3!
58
 6435  x 28  256 x16
 1647360x 44
3
 220  10077696 x 9  
26.
t9 15 C8 x y
7
1
1
1
 x8 , x10  6  x 4
x
x3
 1 
 220(6 x)9   3 
 4x 
29.
1
 1
 2   20 ,
x
x 
 x56
2 1
7
so the x’s go down by 4. So the 4th term will be the
(n  2)!(n  1)n!
n!n(n  1)( n  2)!
19.
4 14
the 9th term will be the
1
6
20
(4calc) and 8 others
9! x 4! = 8709120  total = 3657830400
CCCCTTTOOO
2 2
1
 145152x
18.
x   x 
4 13
1
b)
1
x   x   x
( x 4 )15  x 60 ,
 6435( x 4 )7 (2 x 2 )8
3
14
2 8
so
P4  840
35.
1
2
14
t10  x 5 , so n  k  5
k  9 n  9  5 n  14
17.
7
1
2
3
repeats)
2!
total: 1206
 84  64 x12  27 x 3
a)
b)
c)
d)
x 4! = 360
case 3: no repeats EWDNSAY
C3  3!6 C3  3!2  14400
14 + 28 + 20 + 7
= 69
(n  1)
n(n  1)
14.
5
24.
8
x9
15
13.
+
5 C3 6 C3  200
or
 84(2 x 2 )6 (3x)3
12.
2  16 C2  30
P  P  2  14400
 6435  x 28  256 x16
 1647360x 44
11.
4!
6
2!2!
case 2: EEWDNSAY or DDWENSAY
5 3 6 3
 6435( x ) (2 x )
6
4 CASES
case 1: EEDD
now arrange them around a circle. (6-1)!
Total = 24000
f) BGBGBG x 2
t9 15 C8 x 7 y 8
7! = 5040
f)
7!
 17640
2!
total = 80
So the 9th term will be the
10.
7!
 3! 15120
2!
7! 3!
  7560
2! 2!
4! 6!
  4320
2! 2!
14 C3 5 C1  20
3
4 14
4 7
c)
not Jack & Jill & 3B & 1G
x   x   x
 x56
1 7 
1  14 C2 5 C1  30
 34642080
9.
b)
d) Jack & not Jill & 2B & 2G
 220  19683  x 9  
8.
1
e)
C4 5 C4  325
8!
 20160
2!
a)
c) Jack & Jill & 2B & 1G
3
a) 3 x 2 x 3 x 3 = 54
b) 3 x 2 x 1 x 2 = 12
8 x 2 x 9 = 144
11
WEDNESDAY
d)
C1  7 C2  7 C3  7 C4  7 C5 7 C6 7 C7  127
5 C2 6 C2  150
9
7.
B
a)
so the x’s go down by 4.
 2
 220(3 x)   3 
 x 
8
23.
So the 4th term will be the constant term.
9
2
6
b) 15!  12612600
5!3!4!3!
1
x12 , x11  3  x8 ,
x
1
 x4
x6
2
1
use the 8th row of Pascal’s triangle or
415  1073741824
x10 
1
1
22.
n  10
a)
34.
1
21.
n(n  1)( n  2)  720
10  9  8  720
5.
1
9
C2  6
2.
There are 8 different paths
A
g)
h)
i)
C2 4 C2 4 C2 4 C1  9504
52 C5  39 C5  2023203
13 C1  4 C3 12 C1  4 C2  3744
2 C1  2 C1  12 C1  4 C3  192
13 C2 39 C3  712842