Question

A search committee is formed to find a new software engineer.
a) if 100 applicants apply for the job, how many ways are there to select a subset of 9 for a short list?
b) If 6 of the 9 are selected for an interview, how many ways are there to pick the set of people who are interviewed?
c) Based on the interveiw, the committe wil rank the top three candidates and submit the list to their boss who will make the final decision. How many ways are there to select the list from the 6 interviews?

Answer 1

Answer:

(a) 1,902,231,808,400

(b) 84

(c) 20

Step-by-step explanation:

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

${n\choose k}=\frac{n!}{k!\cdot(n-k)!}$

(a)

Compute the number of ways to select 9 applicants from 100 as follows:

${100\choose 9}=\frac{100!}{9!\cdot(100-9)!}$

        $=\frac{100!}{9!\times 91!}\\\\=\frac{100\times 99\times 98\times 97\times 96\times 95\times 94\times 93\times 92\times 91!}{9!\times 91!}\\\\=\frac{100\times 99\times 98\times 97\times 96\times 95\times 94\times 93\times 92}{9!}\\\\=1902231808400$

(b)

Compute the number of ways to select 6 people from 9 as follows:

${9\choose 6}=\frac{9!}{6!\cdot(9-6)!}$

        $=\frac{9!}{6!\times 3!}\\\\=\frac{9\times 8\times 7\times 6!}{6!\times 3!}\\\\=\frac{9\times 8\times 7}{3!}\\\\=84$

(c)

Compute the number of ways to select top 3 candidates from 6 as follows:

${6\choose 3}=\frac{6!}{3!\cdot(6-3)!}$

        $=\frac{6!}{3!\times 3!}\\\\=\frac{6\times 5\times 4\times 3!}{3!\times 3!}\\\\=\frac{6\times 5\times 4}{3!}\\\\=20$