A committee of 5 persons is to be formed out of 6 gents and 4 ladies. In how many ways this can be done, when at most two ladies are included ?

Options

A. 186
B. 168
C. 136
D. 169

Answer

To solve the problem of forming a committee of 5 persons from 6 gents and 4 ladies, with the condition that at most 2 ladies can be included, we can break it down into different cases based on the number of ladies in the committee.

### Step-by-Step Solution:

**Step 1: Define the Cases**
We need to consider three cases based on the number of ladies included in the committee:
1. Case 1: 0 ladies and 5 gents
2. Case 2: 1 lady and 4 gents
3. Case 3: 2 ladies and 3 gents

**Step 2: Calculate Each Case**

**Case 1: 0 Ladies and 5 Gents**
- We need to select all 5 members from the 6 gents.
- The number of ways to choose 5 gents from 6 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
- Thus, the number of ways is:
\[ \binom{6}{5} = 6 \]

**Case 2: 1 Lady and 4 Gents**
- We need to select 1 lady from 4 and 4 gents from 6.
- The number of ways to choose 1 lady from 4 is \( \binom{4}{1} \), and the number of ways to choose 4 gents from 6 is \( \binom{6}{4} \).
- Therefore, the total number of ways for this case is:
\[ \binom{4}{1} \times \binom{6}{4} = 4 \times 15 = 60 \]

**Case 3: 2 Ladies and 3 Gents**
- We need to select 2 ladies from 4 and 3 gents from 6.
- The number of ways to choose 2 ladies from 4 is \( \binom{4}{2} \), and the number of ways to choose 3 gents from 6 is \( \binom{6}{3} \).
- Therefore, the total number of ways for this case is:
\[ \binom{4}{2} \times \binom{6}{3} = 6 \times 20 = 120 \]

**Step 3: Sum All Cases**
Now, we add the number of ways from all three cases to get the total number of ways to form the committee:
\[ \text{Total} = \text{Case 1} + \text{Case 2} + \text{Case 3} = 6 + 60 + 120 = 186 \]

### Final Answer:
Thus, the total number of ways to form the committee is **186**.

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