distributing $7$ distinct biscuits to $4$ children with following conditions

Question

In how many ways can $7$ distinct biscuits be distributed among $4$ (distinguishable) children such that one child receives $1$ biscuit and the other children receive precisely $2$ biscuits each.

My approach is:

For the child that has $1$ biscuit, there are $4$ choices of choosing the child and $7$ choices of choosing the biscuit.

For the second child that has $2$ biscuits, there are $3$ choices of choosing the child and $C(6,2)$ choices for biscuits.

For the third child that has $2$ biscuits, there are $2$ choices of choosing the child and $C(4,2)$ choices for biscuits.

For the final child that has $2$ biscuits, there are $1$ choices of choosing the child and $C(2,2)$ choices for biscuits.

So, my answer is $4 \cdot 7 \cdot 3 \cdot C(6,2) \cdot 2 \cdot C(4,2) \cdot 1 \cdot 1 = 15120$.

However, the answer is $2520$. Have no idea how to get the correct answer.

Answer 1

Let the children be $A, B, C, D$. There are four cases. Either $A$ is the child who receives one biscuit, $B$ is said child and so on. In each case the child who receives one biscuit has $7$ choices and then we distribute the remaining $6$ biscuits among the remaining $3$ children with each receiving $2$ biscuits, which can be done in $\dfrac{6!}{2!2!2!}$ ways. Putting the pieces altogether, there are $$ 4\times 7\times\frac{6!}{2!2!2!}=2520 $$ ways.

Answer 2

First off, select the child that gets one biscuit. As you said, there are $4$ possible choices for the child and $7$ for the biscuit, giving $4 \times 7$ possibilities.

To count the ways to distribute the remaining biscuits to the other children, imagine to clone each child. So, if you name the children $A$, $B$ and $C$, you now have: $$A, A', B, B', C, C'$$

Now, you have to distribute exactly one biscuit to each. This is a permutation of $6$ elements, so the total number is $6!$. But wait: a child and its clone are the same person, so we don't have to distinguish between them. Thus you have to divide this number by $2!$ (for each child), thereby getting $$4 \times 7 \times \frac{6!}{2!2!2!} = 2520$$