Question

A group of 8 friends (5 girls and 3 boys) plans to watch a movie, but they have only 5 tickets. How many different combinations of 5 friends could possibly receive the tickets? A. 13 B. 40 C. 56 D. 64

Answer 1

The given question is related to combinations .

The options in this case are

All five girls go + 4 girls and 1 boy +3 girls , 2 boys+2 girls , three boys

5C5+ (5C4 * 3C1) + (5C3*3C2) + (5C2*3C3)

=1+5*3 + 10*3 +10*1 = 1+15+30+10 = 56

So the correct option is C .

Answer 2

Given that there are 8 friends (5 girls and 3 boys). We have to select only 5 of them. Those 5 can be girl or boy or combination of both because no such restriction is given.

So basically we have to find how many ways we can select 5 from 8 people.

That can be done using combination formula $C(n,r)=\frac{n!}{r!*(n-r)!}$ where we select r from n people.

Here n=8, r=5

so we get:

$C(8,5)=\frac{8!}{5!*(8-5)!}$

$C(8,5)=\frac{8!}{5!*3!}$

$C(8,5)=\frac{8*7*6*5!}{5!*3*2}$

$C(8,5)=\frac{8*7*6}{3*2}$

$C(8,5)=\frac{8*7}{1}$

$C(8,5)=56$

Hence final answer is 56.