Cobinatoric 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?
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

You got a) wrong. Your answer counts in how many ways you can distribute 16 identical toys among four different children.

 

[lladams]

That doesn't help..thanks anyways

 

[Count Iblis]

That doesn't help..thanks anyways

The answer to a) is 4^16, because for each toy you have four children to shoose from to which it will be given to.

b) 16!/(6!^2 2!^2) * 4!/(2!^2)

4!/(2!^2) counts in how many ways you can choose which children should receive the given amounts, the other factor counts how many ways the toys can be distributed once the choice of which children should receive which amounts is made.

I think you now know how to do c)

 

[lladams]

Thank you

 

[soroban]

Hello, lladams!

How many ways are there to distribute 16 different toys among four children

a. Without restrictions?

b.I f two children get 6 toys and two children get 2 toys?

c. With each child getting 4 toys?

I agree with Count Iblis . . .

(a) Without restrictions

For each toy, there are four choices: which child gets it.
With 16 such decisions, there are: \(\displaystyle \,4^{16}\) possible outcomes.


(b) Two children get 6 toys and two get 2 toys

There are: \(\displaystyle \L\begin{pmatrix}16!\\ 6,6,2,2\end{pmatrix}\) ways to partition the toys into 6-6-2-2.

Then there are \(\displaystyle \L\begin{pmatrix}4\\2\end{pmatrix}\) ways to decide which two children get 6 toys.

Answer: \(\displaystyle \L\:\frac{16!}{(6!)^2(2!)^2}\,\cdot\frac{4!}{2!2!}\)


(c) This is a simple partition of 16 objects into 4-4-4-4:

. . \(\displaystyle \L\begin{pmatrix}16 \\ 4,4,4,4\end{pmatrix} \;=\;\frac{16!}{(4!)^4}\)