class contains b boys and g girls . if the no. of ways of selecting 3 boys and 2 girls from the class is 168 then b+3g=

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The problem involves combinations, a concept in combinatorics. The number of ways to select 3 boys out of b boys is represented as C(b, 3), and the number of ways to select 2 girls out of g girls is represented as C(g, 2). The total number of ways to select 3 boys and 2 girls is given as the product of these two, i.e., C(b, 3)*C(g, 2) = 168.

The formula for combinations is C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and '!' denotes factorial.

So, we have:

b! / [3!(b-3)!] * g! / [2!(g-2)!] = 168.

We know that 3! = 6 and 2! = 2. Also, for large factorials where we only need a part of the product, we can simplify the calculation. For example, b! / (b-3)! = b * (b-1) * (b-2), and similarly g! / (g-2)! = g * (g-1).

So, the equation simplifies to:

[b * (b-1) * (b-2) / 6] * [g * (g-1) / 2] = 168.

This is a nonlinear equation in two variables, b and g. To solve it, we would need another independent equation involving b and g, or we would need to try possible values of b and g that satisfy the equation.

However, the problem does not provide enough information to solve for both b and g. If we had the total number of students, or some other relationship between b and g, we could solve the problem. As it stands, the problem does not have a unique solution.

As for the expression b + 3g, without specific values for b and g, we cannot evaluate it.