A committee of 11 members is to be formed from 8 males and 5 m=females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with atleast 3 females, then

Options

A. `m=n=68`
B. `m+n=68`
C. `m=n=78`
D. `n=m-8`

Answer

To solve the problem, we need to calculate the number of ways to form a committee of 11 members from 8 males and 5 females under two different conditions:

1. **Condition for m**: The committee must have at least 6 males.
2. **Condition for n**: The committee must have at least 3 females.

### Step-by-Step Solution:

**Step 1: Calculate m (at least 6 males)**

We can break this down into three cases based on the number of males:

- **Case 1**: 6 males and 5 females
- **Case 2**: 7 males and 4 females
- **Case 3**: 8 males and 3 females

Now, we will calculate the number of ways for each case.

**Case 1**: 6 males and 5 females
- The number of ways to choose 6 males from 8 is given by \( \binom{8}{6} \).
- The number of ways to choose 5 females from 5 is given by \( \binom{5}{5} \).

So, the total for this case is:
\[ \binom{8}{6} \times \binom{5}{5} = 28 \times 1 = 28 \]

**Case 2**: 7 males and 4 females
- The number of ways to choose 7 males from 8 is \( \binom{8}{7} \).
- The number of ways to choose 4 females from 5 is \( \binom{5}{4} \).

So, the total for this case is:
\[ \binom{8}{7} \times \binom{5}{4} = 8 \times 5 = 40 \]

**Case 3**: 8 males and 3 females
- The number of ways to choose 8 males from 8 is \( \binom{8}{8} \).
- The number of ways to choose 3 females from 5 is \( \binom{5}{3} \).

So, the total for this case is:
\[ \binom{8}{8} \times \binom{5}{3} = 1 \times 10 = 10 \]

Now, we sum the totals from all cases to find \( m \):
\[ m = 28 + 40 + 10 = 78 \]

**Step 2: Calculate n (at least 3 females)**

We can break this down into three cases based on the number of females:

- **Case 1**: 3 females and 8 males
- **Case 2**: 4 females and 7 males
- **Case 3**: 5 females and 6 males

Now, we will calculate the number of ways for each case.

**Case 1**: 3 females and 8 males
- The number of ways to choose 3 females from 5 is \( \binom{5}{3} \).
- The number of ways to choose 8 males from 8 is \( \binom{8}{8} \).

So, the total for this case is:
\[ \binom{5}{3} \times \binom{8}{8} = 10 \times 1 = 10 \]

**Case 2**: 4 females and 7 males
- The number of ways to choose 4 females from 5 is \( \binom{5}{4} \).
- The number of ways to choose 7 males from 8 is \( \binom{8}{7} \).

So, the total for this case is:
\[ \binom{5}{4} \times \binom{8}{7} = 5 \times 8 = 40 \]

**Case 3**: 5 females and 6 males
- The number of ways to choose 5 females from 5 is \( \binom{5}{5} \).
- The number of ways to choose 6 males from 8 is \( \binom{8}{6} \).

So, the total for this case is:
\[ \binom{5}{5} \times \binom{8}{6} = 1 \times 28 = 28 \]

Now, we sum the totals from all cases to find \( n \):
\[ n = 10 + 40 + 28 = 78 \]

### Conclusion:
Thus, we find that \( m = n = 78 \).