The question reads:

In a class of 8 boys and 6 girls, 7 students are to be selected for a tuition fee scholarship. In how many ways can this be done if there must be exactly 3 boys?

Solution:

We need to select exactly 3 boys from 8 boys and 4 girls (since $7 - 3 = 4$) from 6 girls.

1. Number of ways to select 3 boys out of 8:

$\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$

2. Number of ways to select 4 girls out of 6:

$\binom{6}{4} = \binom{6}{2} = \frac{6!}{4!(6-4)!} = \frac{6 \cdot 5}{2 \cdot 1} = 15$

3. Total number of combinations:

The total number of ways to select 3 boys and 4 girls is:

$\binom{8}{3} \cdot \binom{6}{4} = 56 \cdot 15 = 840$

Final Answer:

There are 840 ways to select 7 students with exactly 3 boys.

Would you like further clarification or a breakdown of the steps? Here are related questions for further understanding:

1. How is the combination formula derived and why is it used here?
2. How do we compute $\binom{n}{r}$ for larger values of $n$ and $r$?
3. What if the condition was at least 3 boys instead of exactly 3?
4. How does the problem change if the class size is different?
5. Can we calculate this using a probability-based approach instead of combinations?

Tip: Always verify whether "order matters" in selection problems. If it does, use permutations instead of combinations!