Understanding Combinations in Committees
In combinatorics, when we need to choose a group of items where the order doesn't matter, we use combinations. The formula for choosing $r$ items from a total of $n$ items is given by:
$$C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$
The Problem
We need to form a committee of 5 people from 6 gentlemen and 4 ladies such that there are at least two gentlemen.
Step-by-Step Solution
To solve "at least" problems, we can either sum the cases that satisfy the condition or subtract the unwanted cases from the total.
Case 1: Direct Summation We need 5 people, at least 2 are men (G). The possibilities are:
- 2G and 3L: $\binom{6}{2} \times \binom{4}{3} = 15 \times 4 = 60$
- 3G and 2L: $\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120$
- 4G and 1L: $\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60$
- 5G and 0L: $\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6$
Adding these up: $60 + 120 + 60 + 6 = 246$.
Case 2: Complementary Counting (Faster Method)
- Calculate total ways to choose 5 from 10: $\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
- Subtract the cases that violate the rule (0 or 1 gentleman):
- 0G, 5L: $\binom{6}{0} \times \binom{4}{5} = 0$ (impossible as there are only 4 ladies).
- 1G, 4L: $\binom{6}{1} \times \binom{4}{4} = 6 \times 1 = 6$.
- Total valid ways: $252 - 6 = 246$.
Both methods yield the same result: 246 ways.