More Permutation and Combination questions

nickar1172

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Dec 11, 2013
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Reviewing for my final exam as stated in previous posts so I decided to just post some questions that I have struggled with in the past. If anyone could lend a hand in giving me set ups or formulas for these problems it would be greatly appreciated

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
 
Reviewing for my final exam as stated in previous posts so I decided to just post some questions that I have struggled with in the past. If anyone could lend a hand in giving me set ups or formulas for these problems it would be greatly appreciated

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

I put 11P11 No!

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.....

a) In all circular arrangements of n distinct items there are (n-1)! ways to do it.

b) \(\displaystyle 2\cdot \dbinom{10}{4}\)
 
a) In all circular arrangements of n distinct items there are (n-1)! ways to do it.

b) \(\displaystyle 2\cdot \dbinom{10}{4}\)

Thank you for a) that really helps but I do not understand how you got 2 x 10C4 ?
 
Thank you for a) that really helps but I do not understand how you got 2 x 10C4 ?
How many ways can you select exactly 4 non-defective computers from 10 non-defective computers?

\(\displaystyle \dbinom{10}{4}.\)

How many ways can you select exactly 1 defective computer from 2 defective computers?

\(\displaystyle \dbinom{2}{1} = \dfrac{2!}{1! * (2 - 1)!} = \dfrac{2}{1 * 1} = 2.\)

Same logic in both cases.
 
Thank you for a) that really helps but I do not understand how you got 2 x 10C4 ?

I think that the phrase " identical laptops" is throwing you. In fact no two laptop are identical. (Different serial #s)

There are two with defects and ten with no defects.
To have exactly one in the five with a defect then there are four of the ten without defects.
 
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