Section 7.1
Two formulas are needed for the first 13 problems:
𝑛!
𝑛!
𝑃(𝑛, 𝑘) = (𝑛−𝑘)! and C(n,k) = (𝑛−𝑘)!𝑘!
Later you can use the nPr and nCr functions on your calculators.
Problem 1
8!
C(8,5) = (8−5)!5! =
8!
3! 5!
=
8∗7∗6∗5∗4∗3∗2∗1
3∗2∗1∗5∗4∗3∗2∗1
= 8 ∗ 7 = 𝟓𝟔
Problem 2
8!
P(8,3) = (8−3)! =
8!
5!
=
8∗7∗6∗5∗4∗3∗2∗1
5∗4∗3∗2∗1
= 8 ∗ 7 ∗ 6 = 𝟑𝟑𝟔
Problem 3
8!
P(8,5) = (8−5)! =
8!
3!
=
8∗7∗6∗5∗4∗3∗2∗1
3∗2∗1
= 8 ∗ 7 ∗ 6 ∗ 5 ∗ 4 = 𝟔𝟕𝟐𝟎
Problem 4
6!
6!
C(6,2) = (6−2)!2! = (4)!2! =
6∗5∗4∗3∗2∗1
2∗1∗4∗3∗2∗1
= 3 ∗ 5 = 𝟏𝟓
Problem 5
23!
C(23,19) = (23−19)!19! =
23∗22∗21∗20∗19!
(4)!19!
=
23∗22∗21∗20
4∗3∗2∗1
= 23 ∗ 11 ∗ 7 ∗ 5 = 𝟖𝟖𝟓𝟓
Problem 6
12!
P(12,5) = (12−5)! =
12∗11∗10∗9∗8∗7!
7!
= 12∗ 11 ∗ 10 ∗ 9 ∗ 8 = 𝟗𝟓𝟎𝟒𝟎
Problem 7
11!
P(11,5) = (11−5)! =
11∗10∗9∗8∗7∗6!
6!
= 11 ∗ 10 ∗ 9 ∗ 8 ∗ 7 = 𝟓𝟓𝟒𝟒𝟎
Problem 8
11!
C(11,2) = (11−2)!2! =
11∗10∗9!
(9)!2!
=
11∗10
2∗1
= 11 ∗ 5 = 𝟓𝟓
Problem 9
11!
C(11,5) = (11−5)!5! =
11∗10∗9∗8∗7∗6!
(6)!5!
=
11∗10∗9∗8∗7
5∗4∗3∗2∗1
= 11 ∗ 3 ∗ 2 ∗ 7 = 𝟒𝟔𝟐
Problem 10
5!
P(5,5) = (5−5)! =
5!
0!
= 5*4*3*2*1= 𝟏𝟐𝟎
Problem 11
8!
6!
C(8,4) * C(6,2) = (8−4)!4!*(6−2)!2! =
8∗7∗6∗5∗4!
4!4!
∗
6∗5∗4!
4!2!
=
8∗7∗6∗5
4∗3∗2∗1!
∗
6∗5
2∗1
=
7 ∗ 3 ∗ 5 ∗ 3 ∗ 5 = 𝟏𝟓𝟕𝟓
Problem 12
9!
12!
C(9,6) * C(12,3) = (9−6)!6!*(12−3)!3! =
9∗8∗7∗6!
3!6!
∗
12∗11∗10∗9!
9!3!
=
3 ∗ 4 ∗ 7 ∗ 2 ∗ 11 ∗ 10 = 𝟏𝟖𝟒𝟖𝟎
Problem 13
8!
4!
12!
C(8,2) * C(4,3) / C(12,5) = (8−2)!2!*(4−3)!3!/(12−5)!5! =
8∗7∗6!
6!2!
∗
4∗3!
1!3!
/
12∗11∗10∗9∗8∗7!
=
8∗7
4
12∗11∗10∗9∗8
∗ /
=
1
5∗4∗3∗2∗1
4
𝟏𝟒
4∗7∗
=
𝒐𝒓 𝟎. 𝟏𝟒𝟏𝟒
11 ∗ 9 ∗ 8 𝟗𝟗
7!5!
2∗1
9∗8∗7
3∗2∗1
∗
12∗11∗10
3∗2∗1
=
Problem 14
This is an exercise of using combinations where C(n,k) is the answer to the
given “spot” on the table, with the exception of the values under “Total #
Subsets” (that is the values added across….remember that any time k=0 the
value for the combination is 1 by definition (similar to when any number
raised to the power of zero is equal to 1).
So the table looks like: (which is just Pascal’s triangle)
Set Size,
n (below)
0
1
2
3
4
5
6
7
0
1
1
1
1
1
1
1
1
Number of Subsets of Size k in Set of Size n
Subset Sizes, k
1
2
3
4
5
6
0
0
0
0
0
0
1
0
0
0
0
0
2
1
0
0
0
0
3
3
1
0
0
0
4
6
4
1
0
0
5
10
10
5
1
0
6
15
20
15
6
1
7
21
35
35
21
7
7
0
0
0
0
0
0
0
1
If you note the totals they are all factors of 2 or 2n. So size 10 is 210 or 1024;
size 1 is 21 or 2; and size 50 is 250 or 1.1259x1015.
Problem 15
The first question to ask is whether order matters…in this case “no” because
it does not matter what order the vegetables are placed on the plate, they are
there…so we use combinations. Next, out of the eight choices of vegetables
we must choose 3: C(8,3) = 56
Problem 16
The first question to ask is whether order matters…in this case “no” because
it does not matter what order the members are chosen to teach…so we use
combinations. Next, out of the 12 choices of faculty members we must
choose 5: C(12,5) = 792
Total #
Subsets
1
2
4
8
16
32
64
128
Problem 17
The first question to ask is whether order matters…in this case “yes”
because it different to teach section 1, than to teach section 2…so we will
use permutations. It says “The instructors in Exercise 16 must then be
assigned…” so, Next, out of the 5 chosen faculty members from problem
16, you must now assign them to 5 sections of finite math: P(5,5) = 120
Problem 18
The first question to ask is whether order matters… in this case “no” because
it does not matter what order you chose the hats or the shirts…so we use
combinations. Next, out of the 7 colors of shirts we must choose 3 AND out
of the 5 colors of hats we must choose 2: C(7,3)*C(5,2) = 350
Problem 19
The first question to ask is whether order matters… in this case “no” because
it does not matter what order you chose the men or women to form a
committee…so we use combinations. Next, out of the 7 men we must
choose 2 AND out of the 5 women we must choose 3: C(7,2)*C(5,3) = 210
Problem 20
The first question to ask is whether order matters…in this case both “yes”
and “no” because it does not matter what order you chose to form a
committee, but it does matter when choosing the three officers (Pres., VP
and Sec.)…so we use combinations for the committee and permutations for
the officers. Next, out of the 12 we must choose 3 officers AND out of the
remaining 9 people we must choose 4 people for the committee:
P(12,3)*C(9,4) = 166320
Problems 21-27
The first question to ask is whether order matters… in this case “no” because
it does not matter what order the sample for the inspector…so we use
combinations. Next, we must recognize that out of the total of 24 coffee
makers, 5 are defective, which means that the other 19 are good coffee
makers…so now the specific questions…
Problem 21
Next, out of the all 24 (because there is no call for whether the 3 chosen
have to be good or defective) we must choose 3: C(24,3) = 2024
Problem 22
Next, out of the 19 good coffee makers (because we only want good ones)
we must choose 3: C(19,3) = 969
Problem 23
Next, out of the 19 good coffee makers we choose 2 AND out of the 5
defective coffee makers we must choose 1: C(19,2)*C(5,1) = 855
Problem 24
Next, out of the 19 good coffee makers we choose 1 AND out of the 5
defective coffee makers we must choose 2 (remember that the inspector
must choose three…so if only one is good that means that the remaining 2
must be defective): C(19,1)*C(5,2) = 190
Problem 25
Next, out of the 5 defective coffee makers (because we only want defective
ones…no good ones) we must choose 3: C(5,3) = 10
Problem 26
If you add up the answers for problems 22-25 they will total to the answer in
problem 21. This is because problem 21 asked you all the possible ways to
choose three without respect to being good or defective; while problems 22
through 25 show all the possible combinations of good and defective that are
possible…so they better match up.
Problem 27
Probability is the number of ways to do something (in this case, 2 good and
1 defective…which is answered in problem 23) divided by the total number
of ways to choose three (which is answered in problem 21):
855/2024 = 0.4224
Problem 28
Choosing one or more means everything but choosing NONE (and there is
only one way to choose none …C(6,0)=1)…so the total number of ways
that you can choose is simply 2 to that power (remember problem 14), so the
answer is: 26 - 1 = 63
Problem 29
Choosing three or more means everything but choosing NONE, ONE or
TWO (C(12,0)+C(12,1)+C(12,2) = 79)…so the total number of ways that
you can choose is simply 2 to that power (remember problem 14), so the
answer is: 212 - 79 = 4017
Problems 30-35
The first question to ask is whether order matters… in this case “no” because
it does not matter what order you are dealt the cards…so we use
combinations. Next, we must recognize what is in a deck of cards…4 suits
(13 cards each), 13 denominations (4 cards each), 12 face cards (4 Kings, 4
Queens and 4 Jacks), etc.…so now the specific questions…
Problem 30
Next, out of the 4 Aces in the deck we must choose all 4, AND since this is a
5-card hand we must choose one other card from the remaining 48 cards in
the deck: C(4,4)*C(48,1) = 48
Problem 31
Next, “at least three kings” means 3 or more kings, so since there are only
four kings in the deck, “at least three kings” means 3 kings OR 4 kings, so
we must calculate both 3 kings and 4 kings and add the results together.
So for 4 kings, out of the 4 Kings in the deck we must choose all 4, AND
since this is a 5-card hand we must choose one other card from the
remaining 48 cards in the deck: C(4,4)*C(48,1) = 48
So for 3 kings, out of the 4 Kings in the deck we must choose 3 kings, AND
since this is a 5-card hand we must choose two other cards from the
remaining 48 cards in the deck: C(4,3)*C(48,2) = 4512
So the answer is 48+4512 = 4560
Problem 32
Next, “at most two face cards” means 2 or fewer face cards, so “at most two
face cards” means 2 face cards OR 1 face card OR 0 face cards (no face
cards at all), so we must calculate 2 face cards, 1 face card and no face cards
and add the results together.
So for 2 face cards, out of the 12 face cards in the deck we must choose 2,
AND since this is a 5-card hand we must choose three other non-face cards
from the remaining 40 cards in the deck: C(12,2)*C(40,3) = 652080
So for 1 face card, out of the 12 face cards in the deck we must choose 1,
AND since this is a 5-card hand we must choose four other non-face cards
from the remaining 40 cards in the deck: C(12,1)*C(40,4) = 1096680
So for 0 face cards, since this is a 5-card hand we must choose five other
non-face cards from the remaining 40 cards in the deck: C(12,0)*C(40,5) =
658008
So the answer is 652080+1096680+658008 = 2406768
Problem 33
Next, out of the 4 Aces in the deck we must choose 2, AND out of the 4
Kings in the deck we must choose 2, AND since this is a 5-card hand we
must choose one other card from the remaining 44 cards (remember it
cannot be a king or an ace) in the deck: C(4,2)*C(4,2)*C(44,1) = 1584
Problem 34
Next, out of the 12 face cards in the deck we must choose 4, AND since this
is a 5-card hand we must choose one other card from the remaining 40 cards
in the deck: C(12,4)*C(40,1) = 19800
Problem 35
Next, out of the 4 Fives in the deck we must choose 3, AND out of the 4
Tens in the deck we must choose 2, (and that is five cards):
C(4,3)*C(4,2) = 24
Problems 36-39
This problem is a two step process: choosing the letters (which can be done
using combinations) and then arranging those 5 letters into a “sequence”.
Since these must both be accomplished you multiply the two steps together
(just like in FCP).…so now the specific questions…
Problem 36
First, it must contain a “b” so we must choose the 1 “b”, AND since we need
3 consonants we must choose 2 additional consonants from the remaining 7
consonants, AND from the 4 vowels we must choose 2. Now we must put
them in order, and since there is no restriction to the order you can place any
of the 5 in the first slot, then any of the remaining 4 in the next slot, and so
forth: C(1,1)*C(7,2)*C(4,2)* 5*4*3*2*1 = 15120
Problem 37
First, it must contain a “b” (since if the sequence must begin with a “b” there
must be a “b” to place there) so we must choose the 1 “b”, AND since we
need 3 consonants we must choose 2 additional consonants from the
remaining 7 consonants, AND from the 4 vowels we must choose 2. Now
we must put them in order, and since there is no restriction to the order you
can place only the “b” in the first slot, then any of the remaining 4 in the
next slot, and so forth: C(1,1)*C(7,2)*C(4,2)* 1*4*3*2*1 = 3024
Problem 38
First, it must contain an “e” (since if the sequence must begin with a “e”
there must be an “e” to place there) so we must choose the 1 “e”, AND since
we need 2 vowels we must choose 1 additional vowel from the 3 vowels,
AND from the 8 consonants must choose 3. Now we must put them in
order, and since there is no restriction to the order you can place only the “e”
in the first slot, then any of the remaining 4 in the next slot, and so forth:
C(1,1)*C(3,1)*C(8,3)* 1*4*3*2*1 = 4032
Problem 39
First, it must contain an “e” (since if the sequence must begin with a “e”
there must be an “e” to place there) so we must choose the 1 “e”, AND since
we need 2 vowels we must choose 1 additional vowel from the 3 vowels,
AND it must contain a “b” (since if the sequence must begin with a “b” there
must be a “b” to place there) so we must choose the 1 “b”, AND since we
need 3 consonants we must choose 2 additional consonants from the
remaining 7 consonants. Now we must put them in order, and since there is
no restriction to the order you can place only the “e” in the first slot, then
any of the remaining 4 in the next slot, and so forth:
C(1,1)*C(3,1)*C(1,1)*C(7,2)* 1*4*3*2*1 = 1512
Problems 40-44
The first question to ask is whether order matters… in this case “no” because
it does not matter what order you are chosen to be part of the committee…so
we use combinations. Next, we must recognize that of the total class of 12
there are 9 men and 3 women.…so now the specific questions…
Problem 40
Next, out of the class of 12 we must choose 4, since there is no destinction
between male and female in this question: C(12,4) = 495
Problem 41
Next, out of the 3 women we must choose 1, and since it is a committee of
4, that means that the other 3 must be chosen from the 9 men:
C(3,1)*C(9,3) = 252
Problem 42
Next, since there are no women, that means all 4 are men out of the total of 9
men: C(9,4) = 126
Problem 43
Next, “more men than women” means that in a committee of 4, I could have
all 4 men (zero women), OR 3 men and one woman…anything else would
give equal women to men or more women than men…so we must calculate
both possible outcomes and add the results…
For all men that is answered in problem 42, while 3 men and a woman, is
answered in problem 41…giving an answer of: 126+252 = 378
Problem 44
Next, “at least one women” means that one or more women in the committee
of 4, or every way except all 4 men (zero women), so the total possible
results (problem 40) less all men (problem 42) …giving an answer of:
495-126 = 369
Problem 45
Probability is the number of ways to do something (in this case, one
woman…which is answered in problem 41) divided by the total number of
ways to choose a committee of 4 (which is answered in problem 40):
252/495 = 0.5091
Problem 46
Probability is the number of ways to do something (in this case, at least one
woman…which is answered in problem 44) divided by the total number of
ways to choose a committee of 4 (which is answered in problem 40):
369/495 = 0.7455
Problem 47
Probability is the number of ways to do something (in this case, no
women…which is answered in problem 42) divided by the total number of
ways to choose a committee of 4 (which is answered in problem 40):
126/495 = 0.2545
Problem 48
Probability is the number of ways to do something (in this case, more men
…which is answered in problem 43) divided by the total number of ways to
choose a committee of 4 (which is answered in problem 40):
378/495 = 0.7636
Problems 49-56
The first question to ask is whether order matters… in this case “no” because
it does not matter what order you are dealt the cards…so we use
combinations. Next, we must recognize what is in a deck of cards…4 suits
(13 cards each), 13 denominations (4 cards each), 12 face cards (4 Kings, 4
Queens and 4 Jacks), etc.…so now the specific questions…
Problem 49
Next, out of the 13 Spades in the deck we must choose all 5 cards:
C(13,5) = 1287
Problem 50
Since a suit is not given to us for the flush, we must first choose one of the
four suits, and then, out of the 13 Spades in the deck we must choose all 5
cards: C(4,1)*C(13,5) = 5148
Problem 51
Since it is an “ace-high spade flush”, it must contain the Ace of Spades (of
which there is only one in the deck…next, out of the remaining 12 Spades in
the deck we must choose the other 4 cards:
C(1,1)*C(12,4) = 495
Problem 52
Since a neither the pair or the three of a kind has a denomination called out
we must choose them…you can start with either the pair or the three of a
kind (it does not matter)…so you choose one of the denominations from the
13 that are possible AND pick a pair of that denomination, then pick another
denomination from the remaining 12 AND then pick 3 of the four cards from
that denomination: C(13,1)*C(4,2)*C(12,1)*C(4,3) = 3744
Problem 53
Since a the pair has a no denomination called out, but the three of a kind
does, we must choose the denomination for the pair out of the remaining 12
possible denominations…similar to problem 52:
C(4,3)*C(12,1)*C(4,2) = 288
Problem 54
For the Aces pick 3 of the four aces…now there are two cards which cannot
match (otherwise they would be a pair), so they are from two different
denominations from the remaining 12 AND then from each of those two
denominations you must choose 1 of those 4 cards:
C(4,3)*C(12,2)*C(4,1)*C(4,1) = 4224
Problem 55
Since a none of the cards has a denomination called out we must choose
them… so you choose one of the denominations from the 13 that are
possible AND pick 3 of the four cards from that denomination…now there
are two cards which cannot match (otherwise they would be a pair), so they
are from two different denominations from the remaining 12 AND then from
each of those two denominations you must choose 1 of those 4 cards:
C(13,1)*C(4,3)*C(12,2)*C(4,1)*C(4,1) = 54912
Problem 56
Since a none of the cards has a denomination called out we must choose
them… so you choose two of the denominations from the 13 that are
possible AND pick 2 of the four cards from each denomination (the reason
you do not choose one of the 13 and then one of the remaining twelve, is that
you will get duplication)…now you must choose 1 card from the remaining
44 cards: C(13,2)*C(4,2)* C(4,2)*C(44,1) = 123552
Problems 57-64
Turning a “how many” into a probability is a matter of dividing by the total
number of ways…in this case the total number of ways of choosing 5 cards
out of 52…C(52,5) = 2598960…so now to the specific problems
Problem 57
Probability: 1287/2598960 = 0.000495
Problem 58
Probability: 5148/2598960 = 0.001981
Problem 59
Probability: 495/2598960 = 0.000190
Problem 60
Probability: 3744/2598960 = 0.001441
Problem 61
Probability: 288/2598960 = 0.000111
Problem 62
Probability: 4224/2598960 = 0.001626
Problem 63
Probability: 54912/2598960 = 0.021129
Problem 64
Probability: 123552/2598960 = 0.047539
Problem 65
We know from the earlier chapters that all the ways to line-up7 people is 7!
(or 5040), and of these they will either sit together or they will not. We also
know from earlier that the number of ways to sit together would be 2*6! (or
1440), so the remaining ways are all the ways to create a line-up so that two
people do not sit together…5040-1440 = 3600
Problem 66
First you must choose one of the specific three for the front seat and then
any of the remaining 4 can be arranged in the remaining four seat:
C(3,1)*4! = 72
Problem 67
Turning the number of how many ways in problem 66 into a probability
must be divided by the total number of ways to fill the 5 seats (5! = 120):
72/120 = 0.60