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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?

User Danem
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1 Answer

6 votes

Answer:

There are 27,720 ways to select the committee

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=(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=(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=(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

User Mlienau
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