Suppose a class comprises eight girls and six boys. How many ways can a committee of three students of among 8 girls and 6 boys be formed if the committee must include at least one girl?
My solution:
Ways to form a committee with exactly $g$ girls:
$g = 1$: $\displaystyle\binom 8 1 \binom 6 2 = 120$
$g = 2$: $\displaystyle\binom 8 2 \binom 6 1 = 168$
$g = 3$: $\displaystyle\binom 8 3 = 56$
I have no idea what to do next.
Best Answer
I will assume from your work that there are $\bf 8$ girls AND $\bf 6$ boys (I will take your "9" in the second computation to be a typo), and you need to determine how many committees can be formed from those $14$ students that have at least one girl on the committee of three.
The "simpler approach" to this would be to take the total number of possible ways of selecting any committee of 3 students from among $ 8 + 6 = 14$ students, and then subtract from that total the number of committees/combinations that can be formed without girls. The result will give you the number of committees which then contain at least one girl.
$$\binom{14}{3} - \binom 63 = \text{total number of committees of 3 students with $\bf at\;least\; one\; girl$.}$$
Alternatively, you can add the results of your posted calculations, and arrive at the same number of valid combinations as you would using this approach. Both approaches give you a total of $344$ possible valid combinations.