Math Final review

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nickar1172

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Can somebody let me know if these answers are correct or what do I need to do in order to solve them..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
 
nickar1172 said:
Can somebody let me know if these answers are correct or what do I need to do in order to solve them..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
Click to expand...
You got (a) basically correct: \(\displaystyle \dbinom{4}{1} * \dbinom{9}{3} * \dbinom{7}{2}= \dfrac{4!}{1! * 3!} * \dfrac{9!}{3! * 6!} * \dfrac{7!}{2! * 5!} =\dfrac{4}{1} * \dfrac{9 * 8 * 7}{3 * 2} * \dfrac{7 * 6}{2} = 4 * 84 * 21 = 7056.\)

But (b) follows the same logic: \(\displaystyle \dbinom{2}{1} * \dbinom{6}{1} * \dbinom{3}{1}= \dfrac{2!}{1! * 1!} * \dfrac{6!}{1! * 5!} * \dfrac{3!}{1! * 2!} =\dfrac{2}{1} * \dfrac{6}{1} * \dfrac{3}{1} = 2 * 6 * 3 = 36.\)
 
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