Mathematics, 11.07.2019 05:00, laceybabin1952

Acompany must select 4 candidates to interview from a list of 12, which consist of 8 men and 4 women. how many selections of 4 candidates regardless of gender? how many selections are possible, if 2 women must be selected? how many selections are possible, if at least 2 women must be selected?

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Part A

If 4 candidates were to be selected regardless of gender, that means that 4 candidates is to be selected from 12.

The number of possible selections of 4 candidates from 12 is given by

$^{12}C_4= \frac{12!}{4!(12-4)!}= \frac{12!}{4!\times8!} =11\times5\times9=495$

Therefore, the number of selections of 4 candidates regardless of gender is 495.

Part B:

If 4 candidates were to be selected such that 2 women must be selected, that means that 2 men candidates is to be selected from 8 and 2 women candidates is to be selected from 4.

The number of possible selections of 2 men candidates from 8 and 2 women candidates from 4 is given by

$^{8}C_2\times ^{4}C_2= \frac{8!}{2!(8-2)!}\times 
\frac{4!}{2!(4-2)!} \\ \\ = 
\frac{8!}{2!\times6!}\times\frac{4!}{2!\times2!} 
=4\times7\times2\times3=168$

Therefore, the number of selections of 4 candidates such that 2 women must be selected is 168.

Part 3:

If 4 candidates were to be selected such that at least 2 women must be selected, that means that 2 men candidates is to be selected from 8 and 2 women candidates is to be selected from 4 or 1 man candidates is to be selected from 8 and 3 women candidates is to be selected from 4 of no man candidates is to be selected from 8 and 4 women candidates is to be selected from 4.

The number of possible selections of 2 men candidates from 8 and 2 women candidates from 4 of 1 man candidates from 8 and 3 women candidates from 4 of no man candidates from 8 and 4 women candidates from 4 is given by

$^{8}C_2\times ^{4}C_2+ ^{8}C_1\times ^{4}C_3+ ^{8}C_0\times ^{4}C_4 \\ \\ = \frac{8!}{2!(8-2)!}\times \frac{4!}{2!(4-2)!}+\frac{8!}{1!(8-1)!}\times \frac{4!}{3!(4-3)!}+\frac{8!}{0!(8-0)!}\times \frac{4!}{4!(4-4)!} \\ \\ = \frac{8!}{2!\times6!}\times\frac{4!}{2!\times2!}+\frac{8!}{1!\times7!}\times\frac{4!}{3!\times1!}+\frac{8!}{0!\times8!}\times\frac{4!}{4!\times0!} \\ \\ =4\times7\times2\times3+8\times4+1\times1=168+32+1=201$

Therefore, the number of selections of 4 candidates such that at least 2 women must be selected is 201.

answered: jdkrisdaimcc11

Selections of 4 candidates regardless of gender would result in 12c4, or 12!/(8!*4!) = 9*10*11*12 = 495 possibilities. Selections of 4 candidates where exactly two must be women would result in 8c2 (for the men) * 4c2 (for the women) = 8!/(6!*2!) * 4!/(2!*2!) = 28 * 6 = 168 possibilities. Selections of 4 candidates where AT LEAST two must be women would result in the above 168 (2 women) plus groups where there are exactly three women (8c1*4c3 = 8*(4!/(3!*1!) = 8*4 = 32) plus groups where there are exactly 4 women (1). So there are 168 + 32 + 1 = 201 possible selections of 4 from this group where at least 2 are women.

answered: Guest

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