Math final review: counting problems

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Discussion Overview

The discussion revolves around solving counting problems related to combinatorial selections and options in a math final review context. Participants seek validation of their solutions and clarification on the correct approaches to the problems presented.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents two counting problems and provides their solutions, seeking confirmation of correctness.
  • Another participant agrees with the first answer regarding the selection of a manager but points out a potential typo in notation, suggesting it should be "4C1" instead of "1C4".
  • Concerns are raised about the use of factorials in the second problem, with one participant arguing that the correct approach should involve simple multiplication of options rather than factorial calculations.
  • A later reply reiterates the need to drop the factorials and provides an alternative calculation method for the second problem, emphasizing the number of options available for each choice.

Areas of Agreement / Disagreement

There is partial agreement on the first problem's solution, but disagreement exists regarding the second problem's approach, with multiple interpretations of how to calculate the options available.

Contextual Notes

Participants express uncertainty about the correct application of combinatorial principles and the interpretation of the problems, particularly in the context of factorials versus simple multiplication.

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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