Combinatoric Problem

[lladams]

I need help with this problem, if anyone can help...
How many ways are there to distribute 16 different toys among four children?
a. Without restrictions?
I got c(16+4-1, 4-1)... is this right?
b.If two children get 6 toys and two children get 2 toys?
c.With each child getting 4 toys?
-I'm confused on b and c

 

[Count Iblis]

I need help with this problem, if anyone can help...
How many ways are there to distribute 16 different toys among four children?
a. Without restrictions?
I got c(16+4-1, 4-1)... is this right?

It's right if the toys are all identical. If the toys are all different then the answer is 4^(16). For every toy you have four choices for which child it will be given to. All the different choices will lead to different possibilities because each toy is different.

 

[galactus]

I need help with this problem, if anyone can help...
How many ways are there to distribute 16 different toys among four children?
a. Without restrictions?
I got c(16+4-1, 4-1)... is this right?

This would be the case if the toys were identical, but they are different.

\(\displaystyle \L\\4^{16}\) ways.

 

[pka]

To do part (b) the answer is:\(\displaystyle {4 \choose 2}\frac{{\left( {16!} \right)}}{{\left( {4!} \right)^2 \left( {2!} \right)^2 }}\).

To see this, imagine the sixteen toys in a line. Think of the children as Adam, Barb, Casy, and Devin. How many ways are there to arrange 6-A’s, 6-B’s, 2-C’s & 2-D’s in a corresponding line. That is the number of ways for each of Adam and Barb to get six toys while the other two get two each. Now there are \(\displaystyle 4 \choose 2\) ways to choose two children to get the six toys.

Actually part (c) is easier that that. You do that part.