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.
Why is the answer $931$ and not ${6 \choose 1 }\cdot { 13 \choose 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?
A. EightB. NineC. TenD. ElevenE. Twelve
How does this question have nothing to do with factoring but rather something about combinatorics?
Include one positive integer, positive integer has 2 factors
Include the product of two different positive integers , it has 4 factors
Include product of three, that number has 8 factors
Include product of four, that number has 16 factors.
" 2) there are two ways for each number - select or not factors. total ways is $2 \cdot 2 \cdot 2 \cdot 2=16$. Now this includes the $4$ ways when each is taken alone, so subtract $4$ from $16 = 16-4=12$ It also includes the way when none is selected, so subtract $1$. $12-1=11$."
We have to do two or more, so you aren't doing 'each is taken alone.'
I think the answer is $14=16-2$.
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 $$\binom{6}{1}\binom{8}{3} + \color{red}{\binom{2}{1}}\binom{6}{2}\binom{8}{2} + \color{red}{\binom{3}{1}}\binom{6}{3}\binom{8}{1} + \color{red}{\binom{4}{1}}\binom{8}{4}\binom{6}{0} = \color{red}{\binom{6}{1}\binom{13}{3}}$$
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.