Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects is given by the binomial coefficient: ${n \choose r} = \frac{n!}{r!(n-r)!}$
• ${n \choose r} = {}^nC_r = \frac{n(n-1)(n-2)...(n-r+1)}{r!}$

Step-by-Step Solution

Step 1: Set up the equation based on the problem statement.

We are given that the number of ways to select 3 boys from $b$ boys and 2 girls from $g$ girls is 168. Using the combination formula, we can write this as: ${}^b{C_3} \cdot {}^g{C_2} = 168$

Step 2: Expand the combinations using the formula.

We expand the combination terms using the formula ${}^nC_r = \frac{n(n-1)(n-2)...(n-r+1)}{r!}$: $\frac{b(b-1)(b-2)}{3!} \cdot \frac{g(g-1)}{2!} = 168$ $\frac{b(b-1)(b-2)}{6} \cdot \frac{g(g-1)}{2} = 168$

Step 3: Simplify the equation.

Multiply both sides by 12 to get rid of the fractions: $b(b-1)(b-2) \cdot g(g-1) = 168 \cdot 6 \cdot 2$ $b(b-1)(b-2) \cdot g(g-1) = 168 \cdot 12$ $b(b-1)(b-2) \cdot g(g-1) = 2016$

Step 4: Prime factorize 2016.

Prime factorize 2016 to help find suitable values for $b$ and $g$: $2016 = 2^5 \cdot 3^2 \cdot 7 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 7$

Step 5: Express 2016 as a product of two terms, one being the product of three consecutive integers and the other being the product of two consecutive integers.

We can rewrite 2016 as: $2016 = 8 \cdot 7 \cdot 6 \cdot 3 \cdot 2$ So, $b(b-1)(b-2) = 8 \cdot 7 \cdot 6$ and $g(g-1) = 3 \cdot 2$. This gives us $b=8$ and $g=3$.

Step 6: Calculate $b+3g$.

Now, we can find the value of $b+3g$: $b+3g = 8 + 3(3) = 8 + 9 = 17$

However, the problem states that the correct answer is 3. This indicates an error in the problem itself, or the provided "correct answer".

Let's assume the problem is correct and the answer is 3. Then $b+3g = 3$. Since $b$ and $g$ must be positive integers, if $g=1$, $b=0$, which is not possible since we select 3 boys. Thus, there are no positive integer solutions for $b$ and $g$ such that $b+3g = 3$.

Given the constraint that we must arrive at the given "Correct Answer", let's re-examine the possible factorizations of 2016. $2016 = 2^5 \cdot 3^2 \cdot 7$ We need $b(b-1)(b-2) g(g-1) = 2016$. Suppose $g=2$, then $g(g-1) = 2$. So $b(b-1)(b-2) = 1008$. Testing small values, we find if $b=10$, $b(b-1)(b-2) = 10 \cdot 9 \cdot 8 = 720$. If $b=11$, $b(b-1)(b-2) = 11 \cdot 10 \cdot 9 = 990$. If $b=12$, $b(b-1)(b-2) = 12 \cdot 11 \cdot 10 = 1320$. So, no integer solution in this case. Suppose $g=3$, then $g(g-1) = 6$. So $b(b-1)(b-2) = 336$. Testing small values, we find if $b=8$, $b(b-1)(b-2) = 8 \cdot 7 \cdot 6 = 336$. So, $b=8$ and $g=3$. Then $b+3g = 8+3(3) = 17$.

Since we must arrive at the answer 3, let's assume the question intended to ask for something else. Given $b=8$ and $g=3$, let's find what the value of $b+3g$ should be, if the answer is supposed to be 3. If $b+3g = 3$, since $b$ and $g$ must be positive integers, then $g$ can only be 1 and $b=0$, which is not possible.

It is likely there is an error in the problem statement or the provided answer. Given our calculations, the value of $b+3g$ is 17.

Common Mistakes & Tips

• Double-check the arithmetic, especially when simplifying factorials.
• Consider the practical constraints of the problem (e.g., the number of boys and girls must be non-negative integers).
• If the answer obtained doesn't match the given correct answer, review the problem statement and calculations for errors.

Summary

We are given that ${}^b{C_3} \cdot {}^g{C_2} = 168$. Expanding this equation, we get $\frac{b(b-1)(b-2)}{6} \cdot \frac{g(g-1)}{2} = 168$. Simplifying, we have $b(b-1)(b-2) \cdot g(g-1) = 2016$. By factoring 2016 and matching the factors, we find that $b=8$ and $g=3$ is a solution. Therefore, $b+3g = 8+3(3) = 17$. However, the problem states that the answer is 3, which is not possible based on the derived values of b and g.

The final answer is \boxed{17}.