If one quarter of all the subjects containing three elements of the integers 1,2,3…..m contain the integer 5, then m is equal to

Options

A. 12
B. 10
C. 14
D. 11

Answer

To solve the problem, we need to find the value of \( m \) such that one quarter of all the combinations of selecting three elements from the integers \( 1, 2, 3, \ldots, m \) contains the integer 5.

### Step-by-Step Solution:

1. **Understanding the Total Combinations**:
The total number of ways to choose 3 elements from \( m \) elements is given by the combination formula:
\[ \binom{m}{3} = \frac{m(m-1)(m-2)}{3!} = \frac{m(m-1)(m-2)}{6} \]

2. **Combinations Including the Integer 5**:
If we want to find the number of ways to choose 3 elements that include the integer 5, we can consider 5 as one of the chosen elements. We then need to choose 2 more elements from the remaining \( m - 1 \) integers (i.e., \( 1, 2, 3, \ldots, m \) excluding 5):
\[ \binom{m-1}{2} = \frac{(m-1)(m-2)}{2!} = \frac{(m-1)(m-2)}{2} \]

3. **Setting Up the Equation**:
According to the problem, one quarter of the total combinations must equal the combinations that include the integer 5:
\[ \frac{1}{4} \binom{m}{3} = \binom{m-1}{2} \]

4. **Substituting the Combinations**:
Substitute the expressions we derived into the equation:
\[ \frac{1}{4} \cdot \frac{m(m-1)(m-2)}{6} = \frac{(m-1)(m-2)}{2} \]

5. **Clearing the Fractions**:
Multiply both sides by 24 to eliminate the denominators:
\[ 6m(m-1)(m-2) = 12(m-1)(m-2) \]

6. **Dividing by Common Terms**:
Since \( (m-1)(m-2) \) appears on both sides, we can divide both sides by \( (m-1)(m-2) \) (assuming \( m \neq 1 \) and \( m \neq 2 \)):
\[ 6m = 12 \]

7. **Solving for \( m \)**:
Divide both sides by 6:
\[ m = 2 \]

8. **Verifying the Solution**:
We need to ensure that \( m = 2 \) satisfies the original condition. However, since we need to choose 3 elements, \( m \) must be greater than or equal to 3. Thus, we need to check for higher values of \( m \).

Let's check \( m = 6 \):
- Total combinations: \( \binom{6}{3} = 20 \)
- Combinations including 5: \( \binom{5}{2} = 10 \)
- One quarter of total combinations: \( \frac{20}{4} = 5 \) (not equal)

Continuing this process, we find that \( m = 5 \) gives:
- Total combinations: \( \binom{5}{3} = 10 \)
- Combinations including 5: \( \binom{4}{2} = 6 \)
- One quarter of total combinations: \( \frac{10}{4} = 2.5 \) (not equal)

Finally, checking \( m = 7 \):
- Total combinations: \( \binom{7}{3} = 35 \)
- Combinations including 5: \( \binom{6}{2} = 15 \)
- One quarter of total combinations: \( \frac{35}{4} = 8.75 \) (not equal)

Continuing this process, we find that \( m = 8 \) gives:
- Total combinations: \( \binom{8}{3} = 56 \)
- Combinations including 5: \( \binom{7}{2} = 21 \)
- One quarter of total combinations: \( \frac{56}{4} = 14 \) (not equal)

Eventually, we find that \( m = 10 \) satisfies the condition:
- Total combinations: \( \binom{10}{3} = 120 \)
- Combinations including 5: \( \binom{9}{2} = 36 \)
- One quarter of total combinations: \( \frac{120}{4} = 30 \) (not equal)

Continuing this process, we find that \( m = 12 \) satisfies the condition:
- Total combinations: \( \binom{12}{3} = 220 \)
- Combinations including 5: \( \binom{11}{2} = 55 \)
- One quarter of total combinations: \( \frac{220}{4} = 55 \) (equal)

Thus, the final answer is:
\[ \boxed{12} \]