A committee of $11$ is to be formed from $8$ men and $5$ women. The number of ways $m$ in which at least $6$ men are included is:
Step-by-Step Solution
Key Concept: At least 6 men in an 11-member committee: $(6M, 5W) : \binom{8}{6}\binom{5}{5} = 28$; $(7M, 4W) : \binom{8}{7}\binom{5}{4} = 40$; $(8M, 3W) : \binom{8}{8}\binom{5}{3} = 10$. Total $= 28 + 40 + 10 = 78$.
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Correct Answer: (1)