MHB Math final review: counting problems

AI Thread Summary
The discussion focuses on two math problems related to counting. For the first problem, the correct way to calculate the number of ways to staff positions is using combinations, specifically 4C1 x 9C3 x 7C2, which yields 7056. In the second problem, the calculation should not involve factorials; instead, it should be 2 x 3 x 6, resulting in 36 options for the car purchase. Participants clarify the correct methods and provide guidance on how to approach the problems. The overall emphasis is on correcting misunderstandings in the application of counting principles.
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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