Question

From a group of 5 men and 7 women, five persons are to be selected to form a committee so that at least 3 men are there in the committee. The number of ways this can be done is

• A. 181
• B. 274
• C. 246
• D. None of these

Hint:

In selection problems, break the problem down into cases based on conditions like "at least" and calculate the number of ways for each case.

Answer 1

The Correct Option is C

Step 1: Total committees of 5 from 12 people: \( \binom{12}{5}=792 \).
Step 2: Committees with fewer than 3 men (0, 1, or 2 men): \( \binom{5}{0}\binom{7}{5}+\binom{5}{1}\binom{7}{4}+\binom{5}{2}\binom{7}{3} = 21+175+350=546 \).
\[ \boxed{792 - 546 = 246} \]

Answer 2

A third way to organize this is with Vandermonde's convolution identity, \( \sum_{k} \binom{5}{k}\binom{7}{5-k} = \binom{12}{5} \), which lets us view the three qualifying cases (3, 4, or 5 men) as the complement, within this identity, of the non-qualifying cases (0, 1, or 2 men).

Vandermonde's identity gives \( \binom{5}{0}\binom{7}{5}+\binom{5}{1}\binom{7}{4}+\binom{5}{2}\binom{7}{3}+\binom{5}{3}\binom{7}{2}+\binom{5}{4}\binom{7}{1}+\binom{5}{5}\binom{7}{0} = \binom{12}{5} = 792 \). We already need only the last three terms (3, 4, or 5 men): \[ \binom{5}{3}\binom{7}{2}+\binom{5}{4}\binom{7}{1}+\binom{5}{5}\binom{7}{0} = 10\times21 + 5\times7 + 1\times1 = 210+35+1=246. \]

1. 181: Does not match any grouping of terms from the Vandermonde expansion that corresponds to "at least 3 men," so it is incorrect.
2. 274: Also does not match the required sub-sum of terms, so it is incorrect.
3. 246: Matches the sum of the last three terms of the Vandermonde expansion exactly.
4. None of these: Not applicable, since 246 is obtained directly.

Splitting Vandermonde's identity into the "at least 3 men" terms reproduces the same total as before.

Therefore, the correct answer is 246.