Answer to Question #104668 in Statistics and Probability for Me

Question

There are 15 boys and 19 girls in a room.
a) Find the number of different groups that contain exactly four girls and five boys.
b) How many groups of 14 have an equal number of boys and girls?
c) How many groups of five have more boys than girls?

Answer 1

(a). 4 girls and 5 boys from 15 boys and 19 girls. The are 15 Combination 5 ways of selecting boys and 19 Combination 4 ways of selecting girls.

$C(n,r)=\frac{n!}{r!(n-r)!}$

Ways of selecting girls is given by

$C(19,4)=\frac{19!}{4!×15!}=3876$

Ways of selecting boys is given by

$C(15,5)=\frac{15!}{5!×10!}=3003$

Thus, the number of ways of selecting 4 boys and five girls is $3876×3003=11,639,628$

(b). Ways of selecting 7 boys and 7 girls.

The number of ways of selecting 7 girls is$C(19,7)=\frac{19!}{7!×12!}=50388$

The number of ways of selecting 7 boys is $C(15,7)=\frac{15!}{7!×8!}=6435$

Thus, there are $50388×6435=324246780$ number of groups consisting of 7 boys and 7 girls.

(c) Groups of 5 with more boys than girls.

The number of boys in such a group can be 3 or 4.

Ways of selecting three boys and two girls is given by $\frac{15!}{3!×12!}× \frac{19!}{2!×17!}=455×171=77805$

Ways of selecting 4 boys and a girl is given by

$\frac{15!}{4!×11!}× \frac{19!}{1!×18!}=1365×19=25935$

Thus, there are $77805+25935=103740$ groups of five with more boys.