Number of ways a group of three boys and three girls can be formed from ten boys and five girls if two particular boys cannot be selected together There are ten boys B1, B2, ...., B10 and five girls G1,
G2, ...., G5 in a class. Then the number of ways of
forming a group consisting of three boys and three
girls, if both B1 and B2 together should not be the
members of a group, is__________.
Official Ans. by NTA (1120)
DOUBT:
ATQ if we need both the boys out of group,
can't we just exclude them from group of boys and select from 8 boys and 5 girls?
Making 8c3*5c3 = 560 the answer ?
 A: 
if we need both the boys out of group, can't we just exclude them from group of boys and select from 8 boys and 5 girls?

Yes, but you'll also have to consider the cases where only one of the critical boys is selected. This adds to the number of possible groups:

*

*If $B_1$ is in the group of boys, then for selecting 2 more boys out of the remaining 8 there are $\binom 82$ ways.


*Same if $B_2$ is chosen.


*In neither $B_1$ nor $B_2$ are chosen, there are $\binom{10-2}3$ possible groups.


*There are $\binom 53$ ways to select a group of 3 girls out if a total of 5 girls.
Taking it all together, there are
$$\begin{align}
\underbrace{\left(\binom 82 + \binom 82 + \binom 83\right)}_{\textstyle\text{boys}}\cdot\binom 53
&= (8\cdot 7 / 2 + 8\cdot 7 / 2 + 8\cdot7\cdot 6/6)\cdot(5\cdot4\cdot3/6) \\
&= (8\cdot 7 + 8\cdot7)\cdot 10 \\
&= 2\cdot 56 \cdot 10 = 1120
\end{align}$$
possible groupings.
A: Alternative approach:
$$\left[\binom{10}{3} \times \binom{5}{3}\right] - \left[\binom{8}{1} \times \binom{5}{3}\right] \tag1 $$
$$= [120 \times 10] - [8 \times 10] = 1120.$$
In (1) above, the first term represents the total number of ways of making the selections, without any regard for whether the two boys are selected together.
In (1) above, the second term represents the total number of ways of violating the constraint against the two boys being together, while making the selections.
