In a group of $14$ students, there are $8$ girls and $6$ boys. Determine the number of ways that a committee of $4$ students which has at least $1$ boy can be chosen from the group.
Let's examine two approaches before we address why your suggestion is wrong.
Method 1: We use complementary counting.
There would be $\binom{14}{4}$ ways to select a committee of four students if there were no restrictions. Of these $\binom{8}{4}$ contain only girls. Since the committee must contain at least one boy, we subtract the number of committees with only girls from the total number of committees, which yields $$\binom{14}{4} - \binom{8}{4}$$
Method 2: We count directly.
Observe that the number of ways of selecting exactly $k$ of the six boys and $4 - k$ of the eight girls is $$\binom{6}{k}\binom{8}{4 - k}$$ Since there must be at least one boy on the committee, the number of admissible committees is $$\binom{6}{1}\binom{8}{3} + \binom{6}{2}\binom{8}{2} + \binom{6}{3}\binom{8}{1} + \binom{6}{4}\binom{8}{0}$$
Why is your method wrong?
The set of boys and the set of additional people from which you are selecting are not disjoint, so you count each selection with $k$ boys $k$ times, once for each way you could designate one of those $k$ boys as the boy on the committee. For instance, suppose you choose the boys Adam and Brian and the girls Claire and Denise. You count this selection twice:
$$ \begin{array}{l l l} \text{boy} & \text{additional people}\\ \hline \text{Adam} & \text{Brian, Claire, Denise}\\ \text{Brian} & \text{Adam, Claire, Denise} \end{array} $$
Notice that $$\binom{6}{1}\binom{8}{3} + \color{red}{\binom{2}{1}}\binom{6}{2}\binom{8}{2} + \color{red}{\binom{3}{1}}\binom{6}{3}\binom{8}{1} + \color{red}{\binom{4}{1}}\binom{8}{4}\binom{6}{0} = \color{red}{\binom{6}{1}\binom{13}{3}}$$
How many positive integers can be expressed as a product of two or more of the prime numbers $5, 7, 11,$ and $13$ if no one product is to include the same prime factor more than once?
Method 1: We count directly.
Since a product must contain two or more of the four prime numbers $5, 7, 11, 13$, it can contain $2$, $3$, or $4$ of those numbers. Hence, the number of admissible products is $$\binom{4}{2} + \binom{4}{3} + \binom{4}{4}$$
Method 2: We use complementary counting.
There are $2^4$ possible subsets of four numbers since we must either include or exclude each number from the subset. Thus, if there were no restrictions, we could form $2^4$ products in which at each number in the set $5, 7, 11, 13$ is used at most once. However, each product must contain at least two of these four numbers. Hence, there are $$2^4 - \binom{4}{0} - \binom{4}{1}$$ admissible products.
