A guard of `12` men is formed from a group of `n` soldiers. It is found that `2` particular soldiers `A` and `B` are `3` times as often together on guard as `3` particular soldiers `C, D` & `E`. Then `n` is equal to

Options

A. `28`
B. `27`
C. `32`
D. `36`

Explanation

To solve the problem, we need to determine the value of \( n \) based on the information provided about the guard formations.

### Step 1: Understand the Problem
We have a guard of 12 men formed from a group of \( n \) soldiers. The problem states that soldiers \( A \) and \( B \) are together 3 times as often as soldiers \( C, D, \) and \( E \).

### Step 2: Calculate the Total Ways to Form the Guard
The total number of ways to choose 12 men from \( n \) soldiers is given by the combination formula:
\[ \binom{n}{12} \]

### Step 3: Calculate the Cases for Soldiers A and B
When soldiers \( A \) and \( B \) are together, we can treat them as a single unit or block. Thus, we have:
- 1 block (AB)
- 10 other soldiers (since we need a total of 12)

This gives us a total of 11 units to choose from. The number of ways to choose the remaining 10 soldiers from the remaining \( n-2 \) soldiers (since \( A \) and \( B \) are already chosen) is:
\[ \binom{n-2}{10} \]

### Step 4: Calculate the Cases for Soldiers C, D, and E
For soldiers \( C, D, \) and \( E \), we can treat them as a single block as well. Thus, we have:
- 1 block (CDE)
- 9 other soldiers (to make a total of 12)

This gives us a total of 10 units to choose from. The number of ways to choose the remaining 9 soldiers from the remaining \( n-3 \) soldiers (since \( C, D, \) and \( E \) are already chosen) is:
\[ \binom{n-3}{9} \]

### Step 5: Set Up the Equation
According to the problem, the number of ways in which \( A \) and \( B \) are together is 3 times the number of ways \( C, D, \) and \( E \) are together:
\[ \binom{n-2}{10} = 3 \cdot \binom{n-3}{9} \]

### Step 6: Simplify the Combinations
Using the property of combinations:
\[ \binom{n-2}{10} = \frac{(n-2)!}{10!(n-12)!} \]
\[ \binom{n-3}{9} = \frac{(n-3)!}{9!(n-12)!} \]

Substituting these into the equation gives:
\[ \frac{(n-2)!}{10!(n-12)!} = 3 \cdot \frac{(n-3)!}{9!(n-12)!} \]

### Step 7: Cancel Out Common Terms
We can cancel \( (n-12)! \) from both sides:
\[ \frac{(n-2)!}{10!} = 3 \cdot \frac{(n-3)!}{9!} \]

### Step 8: Further Simplify
We know that:
\[ (n-2)! = (n-2)(n-3)! \]
Substituting this into the equation gives:
\[ \frac{(n-2)(n-3)!}{10!} = 3 \cdot \frac{(n-3)!}{9!} \]

### Step 9: Cancel \( (n-3)! \)
Now we can cancel \( (n-3)! \):
\[ \frac{(n-2)}{10!} = \frac{3}{9!} \]

### Step 10: Solve for \( n \)
Now we can multiply both sides by \( 10! \):
\[ n - 2 = 3 \cdot \frac{10!}{9!} \]
Since \( \frac{10!}{9!} = 10 \):
\[ n - 2 = 30 \]
Thus,
\[ n = 32 \]

### Final Answer
The value of \( n \) is \( 32 \).