Question

Three out of seven students in the cafeteria line are chosen to answer survey questions. How many different combinations of three students are possible?

Answer 1

Answer:

35 possible combinations would be there.

Step-by-step explanation:

We have total of seven students we have to choose three out of those seven

$^{n}C_r=\frac{n!}{r!\cdot (n-r)!}$

n is the total possibilities n =7 in the given case

And r is the chosen ones

Hence, r=3 in the given case.

Substituting the values of n and r in the formula we get

$^{7}C_3$

$\frac{7!}{3!(7-3)!}$

$\frac{7!}{3!\cdot 4!}$

Using n!=n(n-1)----1

$\frac{7\cdot 6\cdot 5\cdot 4!}{3\cdot 2\cdot 1\cdot 4!}$

Cancel the  common factor we get

$35$  possible combinations would be there.