Problem Solving

We originally have 6 people, of which we are trying to choose two - in other words, we have a 6 choose 2 combination:
$$\frac{6!}{\left(6-2\right)!2!}=\frac{6!}{4!2!}=\frac{6\cdot5\cdot4!}{4!\cdot2\cdot1}=\frac{6\cdot5}{2\cdot1}=15$$
So we originally have 15 ways in which the two offers can be allocated.

Then 1/3 of the candidates are disqualified. 1/3 of 6 is 2, so 2 candidates are disqualified and 4 remain. Then, 3 candidates replace them. Now there are 7 candidates to choose from to fill the two job offers. So we now have a 7 choose 2 combination:
$$\frac{7!}{\left(7-2\right)!2!}=\frac{7!}{5!2!}=\frac{7\cdot6\cdot5!}{5!\cdot2\cdot1}=\frac{7\cdot6}{2\cdot1}=21$$
So we now have 21 ways in which the two offers can be allocated.

The answers take the form of percent change, so we'll want to use the percent change equation:
$$\frac{new\ -\ old}{old}\cdot100=\frac{21-15}{15}\cdot100=\frac{6}{15}\cdot100=\frac{2}{5}\cdot100=\frac{200}{5}=40$$
So the number of ways in which the two offers can be allocated increased by 40%, or answer choice D.

swerve wrote:A group of candidates for two analyst positions consists of six people. If one-third of the candidates are disqualified and three new candidates are recruited to replace them, the number of ways in which the two job offers can be allocated will:

A. Drop by 40%
B. Remain unchanged
C. Increase by 20%
D. Increase by 40%
E. Increase by 60%

The number of ways to choose 2 candidates from 6 is 6C2 = 6![2!(6-2)!] = 6!/(2! 4!) = (6 x 5)/2! = 30/2 = 15. After 2 of the 6 candidates leave and 3 new candidates are recruited, there will be 7 candidates, and the number of ways to choose 2 candidates from 7 is 7C2 = 7!/[2!(7-2)!] = 7!/(2!5!) = (7 x 6)/2! = 42/2 = 21. Thus, the number of ways to choose 2 candidates is increased by

(21 - 15)/15 x 100% = 6/15 x 100% = 2/5 x 100% = 40%

Answer: D