Animated Solution for Mathematics - Permutations and Combinations: 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 ______.

Imagine you are standing in a classroom, looking at a group of $b$ boys and $g$ girls. We are tasked with selecting a committee of $3$ boys and $2$ girls, and we are told there are exactly $168$ ways to do this.

We start by translating the physical act of selection into the language of combinations. The number of ways to choose $3$ boys from $b$ is given by the combination formula ${}^b{C}_3$, and the number of ways to choose $2$ girls from $g$ is ${}^g{C}_2$.

Because these two selections are independent parts of the same committee, we multiply them:

Now, let us expand these terms using the definition ${}^n{C}_r=\frac{n!}{r!(n-r)!}$. For our boys, this becomes:

For the girls, it is:

Substituting these into our equation, we get:

Multiplying the denominators, $6\times 2=12$, we can cross-multiply to clear the fraction:

This gives us the product:

Prime Factorization

We have a product of three consecutive integers and two consecutive integers equal to $2016$. To solve for $b$ and $g$, we break $2016$ into its prime factors:

We need to rearrange these factors into a product of three consecutive integers and two consecutive integers. We look for a sequence of three consecutive integers, such as $8,7,6$.

We can manipulate the factors as follows:

Thus, we have:

By comparing our expanded equation $b(b-1)(b-2)\times g(g-1)$ with our factored result $(8\times 7\times 6)\times (3\times 2)$, we can clearly see that $b=8$ and $g=3$.

The final step is to calculate the value requested: $b+3g$. Substituting our values, we get:

The final answer is 17.