Download Mathematics 102 Solutions for HWK 17d

Mathematics 102
Solutions for HWK 17d
Section 6.4 p372
§6.4 p372 Problem 43. Quality Control. In how many ways can a quality-control engineer
select a sample of 3 transistors for testing from a batch of 100 transistors?
Solution. This is exactly like choosing a committee of 3 from a club with 100 members. The
number of possible samples is
C(100, 3) =
100!
100 × 99 × 98
=
= 161, 700.
3!97!
3×2
§6.4 p372 Problem 45.
Television Programming. In how many ways can a televisionprogramming director schedule six different commercials in the six time slots allocated to commercials during a 1-hour program?
Solution. This is like the problem of procrastinating by arranging books on a shelf. One can
either think of allocating commercials or of filling time slots. For instance, one could fill the first
slot by choosing one of the 6 commercials to schedule into that slot, then fill the second slot by
choosing one of the remaining 5 commercials, and so forth. This is like arranging 6 commercials
in 6 slots on a shelf. The number of ways to do this is
P (6, 6) = 6! = 720.
§6.4 p372 Problem 47. Management Decisions. Weaver and Kline, a stock brokerage firm,
has received six inquiries regarding new accounts. In how many ways can these inquiries by directed
to its 12 account executives if each executive handles no more than one inquiry?
Solution. This problem is similar to the previous one, except that there are more executives
than there are inquiries. Think of each inquiry assignment as a slot to be filled with the name of a
different account executive. In other words, one could assign all the inquiries to executives by first
choosing an executive to receive the first inquiry, then an executive to receive the 2nd inquiry, and
so forth, finally choosing an executive to receive the 6th inquiry. Since no executive can receive
more than one inquiry, there are 12 choices for an assignment of the 1st inquiry, then 11 choices
for assigning the 2nd inquiry, and so forth. The total number of possible assignments of the 6
accounts is
P (12, 6) = 12 × 11 × 10 × 9 × 8 × 7 = 665, 280.
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A. Sontag
April 7, 2001
Math 206 HWK 17d Solns contd
6.4 p372
§6.4 p372 Problem 49. Book Displays. At a college library exhibition of faculty publications,
three mathematics books, four social-science books, and three biology books will be displayed on
a shelf. (Assume that none of the books is alike.)
(a) In how many ways can the ten books be arranged on the shelf?
(b) In how many ways can the ten books be arranged on the shelf if books on the same subject
matter are placed together?
Solution.
(a) This is the same book arrangement problem we’ve discussed before, with 10 books to arrange
in all. The total number of arrangements is
P (10, 10) = 10! = 3, 628, 800.
(b) To arrange the books on the shelf, one could first decide an order for the three subjects (for
instance, biology on the left, math in the middle, and social science on the right), then arrange
the math books in their section, arrange the social science books in their section, and arrange the
biology books in theirs. With three types of books to be arranged, there are P (3, 3) = 3! = 6 ways
to decide on an order for the three subjects. Then there are P (3, 3) = 3! ways to arrange the math
books within their section, P (4, 4) = 4! ways to arrange the social science books, and P (3, 3) = 3!
ways to arrange the biology books. So the number of ways to do a complete arrangement would
be
P (3, 3) P (3, 3) P (4, 4) P (3, 3) = 3! 4! 3! 3! = 5184.
§6.4 p372 Problem 65. A five-card poker hand is dealt from a well-shuffled deck of 52 cards.
How many different such hands form a straight flush? (A straight flush consists of 5 cards in
sequence in the same suit, for instance the cards 4♥, 5♥, 6♥, 7♥, 8♥. For such a straight sequence,
an ace may be played as either high or low. In other words there are the possibilities A2345 and
10JQKA.)
Solution. Rather than thinking of being dealt a hand, think of having the 52 cards and laying
out a hand. To lay out a straight flush, one could first decide which suit the flush should come
from, and then choose a straight flush from that suit.
There are 4 possible choices for which suit to use.
How many choices are there for a flush within that suit? The flushes could be counted off in order
as follows:
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A. Sontag
April 7, 2001
Math 206 HWK 17d Solns contd
6.4 p372
1.
A, 2, 3, 4, 5
2.
2, 3, 4, 5, 6
···
6.
6, 7, 8, 9, 10
7.
7, 8, 9, 10, J
···
10.
10, J, Q, K, A
So in all, in a particular suit, there are 10 possible straight flushes. So the total number of straight
flushes, in other words the total number of ways to choose a suit and then choose a straight flush,
is
4 × 10 = 40.
§6.4 p372 Problem 75. Circular Permutations. In how many ways can five TV commentators be seated at a round table for a discussion?
Solution. The 5! possible linear arrangements (for instance, ADCEB) for the 5 commentators
must be sorted according to which ones give the same circular arrangment. For instance ADCEB
and DCEBA would be in the same group because they give the same arrangement around the
table. Each grouping includes 5 linear arrangements (according to which commentator is listed
first in the linear arrangement), so the total number of circular arrangements, or the total number
of circular permutations of the 5 commentators around the table, is
5!
= 4! = 24.
5
More generally, with n commentators to be seated around a circular table there would be (n − 1)!
possible arrangments.
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A. Sontag
April 7, 2001