If 15 boys of different ages are distributed into 3 groups of 4,5, and 6 boys randomly then the probability that three youngest boys are in different groups is

Options

A. `24/91`
B. `71/91`
C. `67/91`
D. `20/91`

Answer

To find the probability that the three youngest boys are in different groups when 15 boys are distributed into three groups of 4, 5, and 6 boys, we can follow these steps:

### Step 1: Determine the total number of ways to distribute the boys
First, we need to calculate the total number of ways to distribute 15 boys into groups of 4, 5, and 6.

The total ways to choose the first group of 4 boys from 15 is given by:
\[ \binom{15}{4} \]
After choosing the first group, we have 11 boys left. The number of ways to choose the second group of 5 boys from these 11 is:
\[ \binom{11}{5} \]
The remaining 6 boys will automatically form the last group, so we have:
\[ \binom{6}{6} = 1 \]
Thus, the total number of ways to distribute the boys is:
\[ \text{Total ways} = \binom{15}{4} \times \binom{11}{5} \times \binom{6}{6} \]

### Step 2: Calculate the number of favorable outcomes
Next, we calculate the number of ways to ensure that the three youngest boys are in different groups.

1. **Assign the three youngest boys to different groups**: We can assign the three youngest boys to the three groups in \(3!\) (factorial of 3) ways, because there are 3 groups.
2. **Distribute the remaining 12 boys**: After placing the three youngest boys, we have 12 boys left to distribute into the groups:
- The first group now has 3 boys (1 from the youngest), so we need to choose 1 more boy from the remaining 12 boys for this group. This can be done in \(\binom{12}{1}\) ways.
- The second group has 5 boys (1 from the youngest), so we need to choose 4 more boys from the remaining 11 boys. This can be done in \(\binom{11}{4}\) ways.
- The last group will automatically have the remaining boys.

Thus, the number of favorable outcomes is:
\[ \text{Favorable outcomes} = 3! \times \binom{12}{1} \times \binom{11}{4} \]

### Step 3: Calculate the probability
The probability \(P\) that the three youngest boys are in different groups is given by the ratio of the number of favorable outcomes to the total number of ways to distribute the boys:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total ways}} = \frac{3! \times \binom{12}{1} \times \binom{11}{4}}{\binom{15}{4} \times \binom{11}{5} \times \binom{6}{6}} \]

### Step 4: Calculate the values
Now we calculate the values of the combinations and factorials:
- \(3! = 6\)
- \(\binom{12}{1} = 12\)
- \(\binom{11}{4} = 330\)
- \(\binom{15}{4} = 1365\)
- \(\binom{11}{5} = 462\)
- \(\binom{6}{6} = 1\)

Substituting these values into the probability formula:
\[ P = \frac{6 \times 12 \times 330}{1365 \times 462 \times 1} \]

Calculating the numerator:
\[ 6 \times 12 \times 330 = 23760 \]

Calculating the denominator:
\[ 1365 \times 462 = 630630 \]

Thus, the probability becomes:
\[ P = \frac{23760}{630630} \]

### Step 5: Simplifying the fraction
To simplify:
\[ P = \frac{24}{81} \quad \text{(after dividing both numerator and denominator by 990)} \]

### Final Answer
The probability that the three youngest boys are in different groups is:
\[ \frac{24}{81} \text{ or } \frac{8}{27} \]