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More Permutation and Combination questions

nickar1172

New member
Dec 11, 2013
20
Reviewing for my final exam can anyone please help access these problems?

a) How many ways can 11 football players stand in a circular huddle?

I put 11P11

b) 12 identical laptops are in the inventory of a dealer, and 2 have hidden defects. If 5 computers to be shipped are selected at random, find the number of ways that exactly 1 defective computer is included in the shipment.

I know I got this completely wrong and don't even want to put up what I had the answer that I got was 144

c)In part b) above, find the probability that exactly 1 defective computer is included in the shipment

once again since I got b wrong.....

d) In setting up a new department, a corporation executive must select a manager from among 4 applicant, 3 clerks from among 9 applicants, and 2 secretaries from among 7 applicants. In how many ways can these positions be staffed

I got 1C4 x 9C3 x 7C2 = 7056

e) A customer can purchase a sports car in either the convertible or hardtop model, in any of 6 colors, and with any of three accessory packages. How many options are open to the purchaser?

I got 2! x 3! x 6! = 8640
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello, nickar1172!

I am assuming the reason no one responded is because there are so many questions posted. We do ask that you post no more than two questions per thread, and it your questions contain multiple related parts, it is best to post only one such problem per thread.

I know is is a bit late for my help to benefit you in time for your exam, but I thought I would answer the questions for future readers of this topic, who may find my response helpful.

a) How many ways can 11 football players stand in a circular huddle?

Your response of:

\(\displaystyle N=\frac{11!}{(11-11)!}=11!=39916800\)

is not quite correct. Generally in circular arrangements, two such arrangements are said to be the same as long as any object in the arrangement has the same left-right neighbors. There are in fact $11!$ linear arrangements of 11 objects, but for illustrative purposes, consider the following three objects and a linear arrangement:

\(\displaystyle A,B,C\)

Now, suppose C moves to the "head" of the line:

\(\displaystyle C,A,B\)

This is a different linear arrangement, but it is the same circular arrangement. Likewise, if $B$ now moves to the head of the line we have:

\(\displaystyle B,C,A\)

So, we may conclude that for every circular arrangement of $n$ objects, there are $n$ distinct linear arrangements, once for each member being at the head of the line. Hence, we conclude that the number $N$ of circular arrangements of $n$ objects is given by the number of linear arrangements $n!$ divided by $n$:

\(\displaystyle N(n)=\frac{n!}{n}=(n-1)!\)

And so for the 11 football players, we find:

\(\displaystyle N(11)=(11-1)!=10!=3628800\)

b) 12 identical laptops are in the inventory of a dealer, and 2 have hidden defects. If 5 computers to be shipped are selected at random, find the number of ways that exactly 1 defective computer is included in the shipment.

Okay, here we need to find the number of ways $N$ to choose 4 of the 10 good computers AND 1 of the 2 defective computers. Hence:

\(\displaystyle N={10 \choose 4}{2 \choose 1}= \frac{10!}{4!6!}\cdot\frac{2!}{1!1!}=\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2}\cdot\frac{2}{1}= 10\cdot3\cdot7\cdot2= 21\cdot20=420\)

c)In part b) above, find the probability that exactly 1 defective computer is included in the shipment

In part b) we found the number of favorable outcomes, and we now simply need to divide this by the total number of outcomes, which is the number of ways to choose 5 computers from 12:

\(\displaystyle P(X)=\frac{420}{{12 \choose 5}}=\frac{420}{792}=\frac{35}{66}\)

d) In setting up a new department, a corporation executive must select a manager from among 4 applicant, 3 clerks from among 9 applicants, and 2 secretaries from among 7 applicants. In how many ways can these positions be staffed?

Here we must compute the number of ways $N$ to select 1 manager from 4 choices AND 3 clerks from 9 choices and 2 secretaries from 7 choices. Hence, we find:

\(\displaystyle N={4 \choose 1}\cdot{9 \choose 3}\cdot{7 \choose 2}=4\cdot84\cdot21=7056\)

This is what you did. :D

e) A customer can purchase a sports car in either the convertible or hardtop model, in any of 6 colors, and with any of three accessory packages. How many options are open to the purchaser?

Here, we want to apply the fundamental counting principle. We simply find the product of the number of choices for each aspect of the car to compute the number of choices $N$ available to the purchaser:

\(\displaystyle N=2\cdot6\cdot3=36\)