Math final review: counting problems

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SUMMARY

The forum discussion focuses on solving combinatorial counting problems related to staffing and purchasing options. For the staffing problem, the correct calculation is 4C1 x 9C3 x 7C2, resulting in 7056 ways to staff the positions. For the car purchasing options, the correct approach is to multiply the number of choices directly: 2 (models) x 3 (accessories) x 6 (colors), yielding 36 options, not using factorials as initially suggested.

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nickar1172
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Can somebody please give me a heads up on if I solved these two problems correctly, I appreciate it, thank you!

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
 
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Re: Math FINAL review

nickar1172 said:
Can somebody please give me a heads up on if I solved these two problems correctly, I appreciate it, thank you!

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

Hi nickar1172, :)

I agree with your first answer, except for one typo I think you made. Instead of "1C4" it should be "4C1". The result seems correct though.

As for the second question, I don't see why the factorials are needed in each space. The car can be "any of 6 colors", not all 6 colors. So there are 6 options for colors, not 6! options as I see it. Same thing for the other two qualities.

What do you get now?
 
Re: Math FINAL review

Jameson said:
Hi nickar1172, :)

I agree with your first answer, except for one typo I think you made. Instead of "1C4" it should be "4C1". The result seems correct though.

As for the second question, I don't see why the factorials are needed in each space. The car can be "any of 6 colors", not all 6 colors. So there are 6 options for colors, not 6! options as I see it. Same thing for the other two qualities.

What do you get now?

I'm lost on e) I got it wrong on the quiz and tried to redo it but I guess that answer is still wrong would you mind posting how to solve / get the solution please? Also what type of question it is
 
Re: Math FINAL review

nickar1172 said:
I'm lost on e) I got it wrong on the quiz and tried to redo it but I guess that answer is still wrong would you mind posting how to solve / get the solution please? Also what type of question it is

You were close, but you need to drop the factorials. The answer is $2 \times 3 \times 6$ as there are 2 options for the first choice, 3 for the second and 6 for the third.
 

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