A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then $$b + 3g$$ is equal to

We need $$\binom{b}{3} \cdot \binom{g}{2} = 168$$.

We factorize: $$168 = 8 \times 21 = 2^3 \times 3 \times 7$$.

Let us try small values. We need $$\binom{b}{3}$$ and $$\binom{g}{2}$$ to be factors of 168.

$$\binom{b}{3}$$: for $$b = 3$$: 1; $$b = 4$$: 4; $$b = 5$$: 10; $$b = 6$$: 20; $$b = 7$$: 35; $$b = 8$$: 56.

$$\binom{g}{2}$$: for $$g = 2$$: 1; $$g = 3$$: 3; $$g = 4$$: 6; $$g = 5$$: 10; $$g = 6$$: 15; $$g = 7$$: 21.

We need the product to be 168:

If $$\binom{b}{3} = 56$$ ($$b = 8$$), then $$\binom{g}{2} = 3$$ ($$g = 3$$). Check: $$56 \times 3 = 168$$. Yes!

If $$\binom{b}{3} = 4$$ ($$b = 4$$), then $$\binom{g}{2} = 42$$, and $$\binom{g}{2} = 42$$ gives $$g(g-1) = 84$$. Since $$9 \times 8 = 72$$ and $$10 \times 9 = 90$$, no integer solution.

If $$\binom{b}{3} = 35$$ ($$b = 7$$), then $$\binom{g}{2} = 168/35 = 4.8$$. Not integer.

If $$\binom{b}{3} = 20$$ ($$b = 6$$), then $$\binom{g}{2} = 168/20 = 8.4$$. Not integer.

If $$\binom{b}{3} = 10$$ ($$b = 5$$), then $$\binom{g}{2} = 168/10 = 16.8$$. Not integer.

So the only valid solution is $$b = 8, g = 3$$.

Therefore $$b + 3g = 8 + 9 = 17$$.

Hence, the correct answer is 17.