The number of ways in which a committee of 7 | vedclass

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(A) Let $T$ be the number of teachers,$F$ be the number of fathers,and $S$ be the number of students. We need to select $7$ members such that $T \ge 1

Vedclass · Accepted Answer

$1170$

Vedclass · Answer

(A) Let $T$ be the number of teachers,$F$ be the number of fathers,and $S$ be the number of students. We need to select $7$ members such that $T \ge 1, F \ge 1, S \ge 1$ and $T > F+S$.
Since $T+F+S = 7$,the condition $T > F+S$ implies $T > 7-T$,so $2T > 7$,which means $T \ge 4$.
Case $1$: $T=4$. Then $F+S=3$. Possible $(F, S)$ pairs are $(1, 2)$ and $(2, 1)$.
Number of ways = $\binom{6}{4} 	imes [\binom{5}{1} 	imes \binom{4}{2} + \binom{5}{2} 	imes \binom{4}{1}] = 15 	imes [5 	imes 6 + 10 	imes 4] = 15 	imes [30 + 40] = 15 	imes 70 = 1050$.
Case $2$: $T=5$. Then $F+S=2$. Possible $(F, S)$ pair is $(1, 1)$.
Number of ways = $\binom{6}{5} 	imes [\binom{5}{1} 	imes \binom{4}{1}] = 6 	imes [5 	imes 4] = 6 	imes 20 = 120$.
Case $3$: $T=6$. Then $F+S=1$. This is not possible since $F \ge 1$ and $S \ge 1$ implies $F+S \ge 2$.
Total ways = $1050 + 120 = 1170$.

The number of ways in which a committee of $7$ members can be formed from $6$ teachers,$5$ fathers,and $4$ students in such a way that at least one from each group is included and teachers form the majority among them,is

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