A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has at least one boy and one girl? In my basic Mathematics textbook of high school, in the Combination chapter, there is a question as follows:
"A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has at least one boy and one girl?", whose answer is given as 
But, if the boys and girls were indistinguishable (i.e., 4 girls are female and 7 boys are male only), then if we proceed with this assumption (as it is not mentioned that they have different identities), we can take 1 girl and 1 boy pre-reserved for the team of 5 members in 4 and 7 ways respectively, and fill up the remaining 3 vacant seats in the team by choosing any 3 persons from the remaining 9 people (i.e., 3 girls + 6 boys) it will lead us to $4×7×{ }^{(3+6)} C_{3}=2352$ number of ways to form the team. Why is this answer wrong?
 A: What was your error?
By designating a particular girl as the girl in the group and a particular boy as the boy in the group, you count each case multiple times.  You count the case in which $b$ boys and $g$ girls are selected $bg$ times.
Suppose the available girls are Abigail, Barbara, Charlotte, and Denise and that the available boys are Eric, Frank, George, Henry, Ian, James, and Kirk.
You count each selection with one girl and four boys four times, once for each way you could designate one of the four boys as the boy you selected.   For instance, you count the selection Abigail, Eric, Frank, George, and Henry four times.
\begin{array}{l l l}
\text{boy} & \text{girl} & \text{additional people} \\ \hline
\text{Eric} & \text{Abigail} & \text{Frank, George, Henry}\\
\text{Frank} & \text{Abigail} & \text{Eric, George, Henry}\\
\text{George} & \text{Abigail} & \text{Eric, Frank, Henry}\\
\text{Henry} & \text{Abigail} & \text{Eric, Frank, George}\\
\end{array}
You count each selection with two girls and three boys six times, once for each of the two ways you could designate one of the girls as the girl who has been selected and once for each of the three ways you could designate one of the three boys as the boy who has been selected.  For instance, you count the selection Barbara, Charlotte, Ian, James, Kirk six times.
\begin{array}{l l l}
\text{boy} & \text{girl} & \text{additional people} \\ \hline
\text{Ian} & \text{Barbara} & \text{Charlotte, James, Kirk}\\
\text{Ian} & \text{Charlotte} & \text{Barbara, James, Kirk}\\
\text{James} & \text{Barbara} & \text{Charlotte, Ian, Kirk}\\
\text{James} & \text{Charlotte} & \text{Barbara, Ian, Kirk}\\
\text{Kirk} & \text{Barbara} & \text{Charlotte, Ian, James}\\
\text{Kirk} & \text{Charlotte} & \text{Barbara, Ian, James}\\
\end{array}
You count each selection with three girls and two boys six times, once for each of the three ways you could designate one of the three girls as the girl who has been selected and once for each of the two ways you could designate one of the two boys as the boy who has been selected.
You count each selection with four girls and one boy four times, once for each of the four ways you could designate one of the four girls as the girl who has been selected.
Notice that
$$\color{red}{4}\binom{4}{1}\binom{7}{4} + \color{red}{6}\binom{4}{2}\binom{7}{3} + \color{red}{6}\binom{4}{3}\binom{7}{2} + \color{red}{4}\binom{4}{4}\binom{7}{1} = \color{red}{2352}$$
By saying that there are ways to select a girl and seven ways to select a boy, you are treating individual girls as distinct and individual boys as distinct.
Alternate Method:  Note that there is a simpler way of solving the problem.  We can subtract the number of ways of selecting only girls or only boys from the total number of ways of selecting five of the $4 + 7 = 11$ available people.  Observe that since there are only four girls, it is not possible to select only girls.  However, without the restriction, we could have selected only boys.  Therefore, the number of ways of selecting five students from seven boys and four girls so that at least one boy and at least one girl is selected is found by subtracting the number of ways of selecting only boys from the number of ways of selecting five of the eleven people, which is
$$\binom{11}{5} - \binom{7}{5} = 441$$
