A class have `B` boys and `G` girls , 3 boys and 2 girls selected at random and number of ways of selecting 3 boys and 2 girls are 168 . Then `B+2G` is equal to

Answer

To solve the problem, we need to find the values of \( B \) (the number of boys) and \( G \) (the number of girls) such that the number of ways to select 3 boys from \( B \) boys and 2 girls from \( G \) girls equals 168. We will then calculate \( B + 2G \).

### Step-by-Step Solution:

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

2. **Setting Up the Equation**:
The total number of ways to select 3 boys and 2 girls is given by:
\[ \binom{B}{3} \times \binom{G}{2} = 168 \]
Substituting the combinations we found:
\[ \frac{B(B-1)(B-2)}{6} \times \frac{G(G-1)}{2} = 168 \]

3. **Simplifying the Equation**:
Multiply both sides by 12 (to eliminate the fractions):
\[ B(B-1)(B-2) \times G(G-1) = 2016 \]

4. **Finding Factor Pairs**:
We need to find integer values for \( B \) and \( G \) such that the product \( B(B-1)(B-2) \) and \( G(G-1) \) equals 2016. We can start by factoring 2016:
\[ 2016 = 2^5 \times 3^2 \times 7 \]

5. **Testing Values for \( B \)**:
We can test different integer values for \( B \) and calculate \( B(B-1)(B-2) \) until we find a suitable value:
- For \( B = 8 \):
\[ 8 \times 7 \times 6 = 336 \]
Now, we need \( G(G-1) \) such that:
\[ 336 \times G(G-1) = 2016 \implies G(G-1) = \frac{2016}{336} = 6 \]
The pairs \( (G, G-1) \) that satisfy this are \( G = 3 \) (since \( 3 \times 2 = 6 \)).

6. **Calculating \( B + 2G \)**:
Now that we have \( B = 8 \) and \( G = 3 \):
\[ B + 2G = 8 + 2 \times 3 = 8 + 6 = 14 \]

### Final Answer:
\[ B + 2G = 14 \]