Question

RSM wants to send four of its 18 Math Challenge teachers to a conference. How many combinations of four teachers include exactly one of either Mrs. Vera or Mr. Jan?

Answer 1

Answer:

1120 combinations of four teachers include exactly one of either Mrs. Vera or Mr. Jan.

Step-by-step explanation:

The order in which the teachers are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

1 from a set of 2(Either Mrs. Vera or Mr. Jan).

3 from a set of 18 - 2 = 16. So

$C_{2,1}C_{16,3} = \frac{2!}{1!1!} \times \frac{16!}{3!13!} = 1120$

1120 combinations of four teachers include exactly one of either Mrs. Vera or Mr. Jan.