There are ten boys B_1,,B_10 and five girls G_1,,G_5 in a class.

Question

There are ten boys $B_1,\ldots,B_{10}$ and five girls $G_1,\ldots,G_5$ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both $B_1$ and $B_2$ together should not be members of a group, is\_\_\_

Answer: 1120

Accepted Answer

Total ways without restriction: $\binom{10}{3}\binom{5}{3}=120\times10=1200$.

Ways with both $B_1$ and $B_2$ included: choose 1 more boy from remaining 8: $\binom{8}{1}=8$, times $\binom{5}{3}=10$: $80$.

$$\text{Required} = 1200-80 = 1120$$

Solutions · 1