Choose TWO Children.......

The number of total choices is $9 \choose 4$ as we are selecting the number of ways to choose $4$ people out of a group of $9$ . Now, the number of ways that we can select $2$ children exactly out of $4$ is then $4 \choose 2$ . The total number of options where two children are chosen must then be multiplied by the number of ways to select the remaining adults to get the full number of combinations for each selection of children. This product becomes ${4 \choose 2} \times {5 \choose 2}$ , divided by the total number of possibilities becomes our answer, $\frac{{4 \choose 2} \times {5 \choose 2}}{9 \choose 4} = \boxed{10/21}$

In case you're not familiar with "choose" notation, the large parentheses signify the number of ways to choose the bottom number of objects from a collection of the top number of objects. The formula is ${n \choose k} = \frac{n!}{(n-k)! \times k!}$ .

The first question to ask is " How many ways can we pick 4 people from a group of 9?' The answer is 9 C 4 = 126 ways. 9 C 4 = 9! /( 9- 4)!(4!)=126

The second part is we need to select 2 children out of the 4 and this equals to 4 C 2= 6 AND for the remaining two, it can be any combination of women and men, which is 5 C 2= 10

Lastly, combine to ensure that when we selct 4, that there are 2 children is (6 * 10) / 126 = 60/126 = 10/21 Therefore, a+b= 31