A committee of 4 person is to, be selected from 5 men and 6 women.
Quantity I: Number of ways the selection can be done to include at least 1 woman.
Quantity II: Number of ways the selection can be done to include at least 1 man.

Options

A. Quantity `IgtQ` uantity II
B. Quantity `IltQ` uantity II
C. Quantity `IleQ` uantity II
D. Quantity `IgeQ` uantity II

Answer

To solve the problem, we need to calculate the number of ways to form a committee of 4 persons from 5 men and 6 women, ensuring that we include at least one woman for Quantity I and at least one man for Quantity II.

### Step 1: Calculate Total Combinations
First, we calculate the total number of ways to select 4 persons from the total of 11 persons (5 men + 6 women).

\[ \text{Total ways} = \binom{11}{4} \]

### Step 2: Calculate Combinations for Quantity I (At least 1 Woman)
To find the number of ways to select a committee that includes at least one woman, we can use the complementary counting method. We first calculate the number of ways to select a committee with no women (i.e., all men).

- **All Men (0 Women)**: We can only select from the 5 men.
\[ \text{Ways to select 4 men} = \binom{5}{4} = 5 \]

Now, we subtract this from the total combinations to find the number of ways to select at least one woman.

\[ \text{Quantity I} = \text{Total ways} - \text{Ways to select 4 men} \]
\[ \text{Quantity I} = \binom{11}{4} - 5 \]

### Step 3: Calculate Combinations for Quantity II (At least 1 Man)
Similarly, for Quantity II, we calculate the number of ways to select a committee that includes at least one man. Again, we use complementary counting.

- **All Women (0 Men)**: We can only select from the 6 women.
\[ \text{Ways to select 4 women} = \binom{6}{4} = 15 \]

Now, we subtract this from the total combinations to find the number of ways to select at least one man.

\[ \text{Quantity II} = \text{Total ways} - \text{Ways to select 4 women} \]
\[ \text{Quantity II} = \binom{11}{4} - 15 \]

### Step 4: Compare Quantity I and Quantity II
Now we need to compare the two quantities:

1. **Calculate Total Ways**:
\[ \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \]

2. **Calculate Quantity I**:
\[ \text{Quantity I} = 330 - 5 = 325 \]

3. **Calculate Quantity II**:
\[ \text{Quantity II} = 330 - 15 = 315 \]

### Conclusion
Now we can conclude:
\[ \text{Quantity I} = 325 \quad \text{and} \quad \text{Quantity II} = 315 \]

Since \( 325 > 315 \), we find that:

**Quantity I is greater than Quantity II.**

### Final Answer
The answer is that Quantity I is greater than Quantity II.

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