I am having trouble solving for this problem. I would like to see a clear way to obtain the solution.
How many ways you can distribute 10 different toys to 4 children if each child receives at least 2 toys?
What are the possible outcomes? One child can receive 4 toys, each other gets 2. Two children can receive 3 toys, the other two get 2. We will handle each case separately.
For the case that one child receives 4 toys and the other two get 2 each, choose the child who receives 4 toys. There are 4 children to choose and you choose 1 who will receive the 4 toys. Then, choose 4 of the 10 toys to give to the first child. Then, choose 2 of the remaining 6 toys to give to the second child, 2 of the remaining 4 to give to the third child, and 2 of the remaining two to give to the last child.
So, for this case, there are
$\underbrace{\binom{4}{1}}_{\begin{matrix}\text{Ch oose kid} \\ \text{to get 4 toys}\end{matrix}} \underbrace{\binom{10}{4}}_{\begin{matrix}\text{Ch oose 4 toys} \\ \text{for first child}\end{matrix}} \underbrace{\binom{6}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for second child}\end{matrix}} \underbrace{\binom{4}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for third child}\end{matrix}} \underbrace{\binom{2}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for last child}\end{matrix}}$
This can be simplified as:
$\dfrac{4!\cdot 10!}{(1!3!)(4!2!2!2!)} = 75,600$
Next, choose two children to get 3 toys. Then choose 3 of the 10 toys for the first child, 3 of the remaining 7 for the second child, 2 of the remaining 4 for the third, and 2 of the remaining 2 for the last:
$\underbrace{\binom{4}{2}}_{\begin{matrix}\text{Ch oose kids} \\ \text{to get 3 toys}\end{matrix}} \underbrace{\binom{10}{3}}_{\begin{matrix}\text{Ch oose 3 toys} \\ \text{for first child}\end{matrix}} \underbrace{\binom{7}{3}}_{\begin{matrix}\text{Cho ose 3 toys} \\ \text{for second child}\end{matrix}} \underbrace{\binom{4}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for third child}\end{matrix}} \underbrace{\binom{2}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for last child}\end{matrix}}$
This can be simplified as:
$\dfrac{4!\cdot 10!}{(2!2!)(3!3!2!2!)} = 151,200$
Adding these together: $75,600+151,200 = 226,800$
Hello, Yeison!
SlipEternal is absolutely correct.
Here is another approach.
How many ways you can distribute 10 different toys to 4 children
if each child receives at least 2 toys?
There are two possible distributions: .$\displaystyle \begin{Bmatrix}[1] & (2,2,2,4) \\ [2] & (2,2,3,3)\end{Bmatrix}$
Case [1]
There are 4 choices for the child to receive four toys.
Then the toys can be distributed in: $\displaystyle {10\choose2,2,2,4} = 18,\!900$ ways.
. . There are: .$\displaystyle 4\cdot18,\!900 \,=\,75,\!600$ ways.
Case [2]
There are $\displaystyle {4\choose2,2} = 6$ choices for the children to receive three toys.
Then the toys can be distributed in: $\displaystyle {10\choose2,2,3,3} = 25,\!200$ ways.
. . There are: .$\displaystyle 6\cdot 25,\!200 \:=\:151,\!200$ ways.
Therefore, there are .$\displaystyle 75,\!600 + 151,\!200 \:=\:226,\!800$ ways.