Homework #3 — Math 173 — Spring ’09
Solutions
(1) 5.2; 12)
a) This is equivalent to distributing 14 distinct balls (i.e., the people) into three distinct
bins such that 3 go into the first bin, 5 into the second, and 6 into the third. The
answer is the multinomial coefficient 14
3,5,6=14!
3!5!6! = 168168.
b) At first glance, it looks like by the logic above that the answer should be the multinomial
coefficient 14
7,7. However, this is too large by a factor of two. To see this, consider
the smaller problem of partitioning 4 people into two groups of two. Person A has
to go with someone else: B, C, or D. Whoever A is matched with, the other pair is
determined. So there are three possibilities. But this is half of 4
2.
The reason is that the groups are unordered. In terms of balls-and-boxes, we have
erased the labels on the boxes. So we must divide by two and the answer is 1
214
7,7=
1716.
We didn’t run into this problem in part a) because all of the groups were of different
sizes and therefore automatically distinguished/unordered among themselves.
(2) 5.2; 16) We know from class that the number of possible poker hands is 52
5= 2598960.
a) First we choose the aces in 4
4= 1 possible ways, then we choose the fifth card of
the hand in 52−4
1= 48 possible ways. By the MR, the answer is (1 ·48)/2598960 ≈
0.000085.
b) This is thirteen times as likely as the answer to part a) since we now have the ad-
ditional step of choosing which of the thirteen possible values occurs (rather than
being restricted to the single value of Ace). So the answer is (13 ·48)/2598960 =
624/2598960 ≈0.00024.
e) Ignoring suits, there are 10 possible sequences of values: A2345, 23456,. . . ,10JQKA.
Regardless of which sequence is chosen, we have 4 possibilities for the suit of each card.
So the answer is (10 ·45)/2598960 = 10240/2598960 ≈0.0039
f) Not having a pair amounts to choosing five of the thirteen values: 13
5. We must
choose a suit for each selection, leading to a probability of (13
5·45)/2598960 =
131220/2598960 ≈0.507.
Note that it is much harder to solve this using the DR. The number of ways of getting
at least one pair is not 134
248
3(13 choices for the value of the pair; 4
2choices for
the suits of the pair; 48
3ways of choosing the remaining three cards from the 52 −4
cards with value differing from the pair). This answer overcounts. For instance, a pair
might be obtained from the 48
3cards.
(3) 5.2; 22) The number of ways of dealing the cards is the multinomial coefficient 52
4,4,...,4
(thirteen 4’s) since each of the thirteen players gets 4 cards. For each suit, there are 13!
ways to assign the cards of that suit to the 13 distinct players. Since there are four suits in
all, the answer is then (13!)4/52
4,4,...,4= (13!)4(4!)13/52! ≈1.6·10−11 (i.e., unlikely).
(4) 5.2; 26) First choose which 10 men in 20
10possible ways. Assuming the women are lined
up already, we then have 10! ways to line the men up relative to the womens’ ordering. So
the answer is 20
10·10! = 20!/10!.
(5) 5.2; 37) We are told that the “middle” portion of our ordering must look like “· · · vccvccv· · ·
” (where “c” stands for consonant and “v” for vowel). Our first step is to choose how many
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