Question

How many ways are there to select a committee to develop a discrete mathematics course at a school if the committee is to consist of 3 faculty members from the mathematics department and 4 from the computer science department, if there are 9 faculty members of the math department and 11 of the CS department?

Answer 1

Answer:

There are 27,720 ways to select the committee

Step-by-step explanation:

First, it is necessary to know how many ways are there to select 3 members, if there are 9 members of the mathematics department. This can be found using the following equation:

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

Where nCk gives as the number of ways in which we can select k elements from a group of n elements. So, replacing n by 9 and k by 3 members, we get:

$9C3=\frac{9!}{3!(9-3)!}=84$

So, there are 84 ways to select 3 members from 9 members of the mathematics department.

At the same way, we can calculate that there are 330 ways to select 4 members from the 11 that belong to the Computer science department as:

$11C4=\frac{11!}{4!(11-4)!}=330$

Finally the total number of ways in which we can form a committee with 3 faculty members from mathematics and 4 from the computer science department is calculated as:

9C3 * 11C4 = 84 * 330 = 27,720