Quantity I : In how many a committee of 4 members with at least 2 women can be formed from 8 men and 4 women ?
Quantity II : How many 3 - digit numbers which are divisible by 3 can be formed from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 such that 3-digit number always ends with an even number ?

Options

A. Quantity I ` lt ` Quantity II
B. Quantity I ` le ` Quantity II
C. Quantity I ` gt ` Quantity II
D. Quantity I ` ge ` Quantity II

Answer

To solve the problem, we will break it down into two parts: Quantity I and Quantity II.

### Quantity I: Forming a Committee of 4 Members with at least 2 Women

We have 8 men and 4 women. We need to form a committee of 4 members with at least 2 women. The possible combinations are:

1. **2 Women and 2 Men**
2. **3 Women and 1 Man**
3. **4 Women**

Let's calculate each case:

#### Case 1: 2 Women and 2 Men
- Number of ways to choose 2 women from 4:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
- Number of ways to choose 2 men from 8:
\[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]
- Total ways for this case:
\[ 6 \times 28 = 168 \]

#### Case 2: 3 Women and 1 Man
- Number of ways to choose 3 women from 4:
\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = 4 \]
- Number of ways to choose 1 man from 8:
\[ \binom{8}{1} = 8 \]
- Total ways for this case:
\[ 4 \times 8 = 32 \]

#### Case 3: 4 Women
- Number of ways to choose 4 women from 4:
\[ \binom{4}{4} = 1 \]
- Number of ways to choose 0 men from 8:
\[ \binom{8}{0} = 1 \]
- Total ways for this case:
\[ 1 \times 1 = 1 \]

#### Total for Quantity I
Now, we sum the total ways from all cases:
\[ 168 + 32 + 1 = 201 \]

### Quantity II: 3-Digit Numbers Divisible by 3 Ending with an Even Number

We need to find 3-digit numbers that are divisible by 3 and end with an even digit. The even digits available are 0, 2, 4, 6, and 8.

1. **Identify the range of 3-digit numbers**: The smallest is 100 and the largest is 999.
2. **Count the total 3-digit numbers divisible by 3**:
- The first 3-digit number divisible by 3 is 102.
- The last 3-digit number divisible by 3 is 999.
- The sequence of 3-digit numbers divisible by 3 can be expressed as:
\[ 102, 105, 108, \ldots, 999 \]
- This is an arithmetic series where:
- First term \(a = 102\)
- Last term \(l = 999\)
- Common difference \(d = 3\)
- The number of terms \(n\) can be found using the formula:
\[ n = \frac{l - a}{d} + 1 = \frac{999 - 102}{3} + 1 = 300 \]

3. **Count the valid 3-digit numbers ending with an even digit**:
- The valid last digits are 0, 2, 4, 6, and 8.
- For each case, we will consider the last digit and find the number of valid combinations for the first two digits.

#### Case 1: Last digit = 0
- The first digit can be from 1 to 9 (9 options).
- The second digit can be from 0 to 9 (10 options).
- Total: \(9 \times 10 = 90\)

#### Case 2: Last digit = 2
- The first digit can be from 1 to 9 (9 options).
- The second digit can be from 0 to 9 (10 options).
- Total: \(9 \times 10 = 90\)

#### Case 3: Last digit = 4
- The first digit can be from 1 to 9 (9 options).
- The second digit can be from 0 to 9 (10 options).
- Total: \(9 \times 10 = 90\)

#### Case 4: Last digit = 6
- The first digit can be from 1 to 9 (9 options).
- The second digit can be from 0 to 9 (10 options).
- Total: \(9 \times 10 = 90\)

#### Case 5: Last digit = 8
- The first digit can be from 1 to 9 (9 options).
- The second digit can be from 0 to 9 (10 options).
- Total: \(9 \times 10 = 90\)

### Total for Quantity II
Now, we sum the total ways from all cases:
\[ 90 + 90 + 90 + 90 + 90 = 450 \]

### Conclusion
- Quantity I = 201
- Quantity II = 450

Now we compare the two quantities:
- Quantity I (201) is less than Quantity II (450).

### Final Answer
**Quantity I < Quantity II**