# A child has six blocks, three of which are red and three of which are green. How many patterns can she make by placing

A child has six blocks, three of which are red and three of which are green. How many patterns can she make by placing them all in a line? If she is given three white blocks, how many total patterns can she make by placing all nine blocks in a line?

I was able to figure out the solutions as

a) $$\binom{6}{3\ 3}$$

b) $$\binom{9}{3\ 3\ 3}$$

But I was wondering what is the alternative way to count these situations? I understand this manner in that it is describing the number of ways my total number of positions can be divided into subgroups of the respective sizes. But how would I organize the cases if I say did not know this mulitnomial formula?

Edit: The explanations I have been getting have been based on the multinomial formula. The motivation for my question came from the fashion in which I saw this form of the solution:

I apologize for the link from elsewhere. I was trying to understand how this person deconstructed the cases.

Thanks

• Imagine you had 6 different blocks. You would have 6! combinations. But now you consider 3 of them to be same so you will have to divide your 6! by 3!3!. Jun 3, 2016 at 19:06

Alternate way to think of the second answer:

Apply multiplication principle:

• Choose the locations of the red blocks: $\binom{9}{3}$ options

• From the remaining locations, choose the locations of the green blocks: $\binom{6}{3}$ options

• From the remaining locations, choose the locations of the white blocks: $\binom{3}{3}$ options

There are then $\binom{9}{3}\binom{6}{3}\binom{3}{3}$ total arrangements.

Note that $\binom{9}{3}\binom{6}{3}\binom{3}{3}=\binom{9}{3,3,3}$

For the harder question it is a all possible permutations of 9 blocks divided by permutations of 3 blocks of each color

$\frac{9!}{3!3!3!}$

because all blocks sharing same color are indistinguishable in a line.

An intuition-type argument:

If they were lettered blocks, {ABCDEF} in the first case, there would be $6!$ possible orderings of the blocks. If you take one of those and paint {ABC} blocks solid red so they are indistinguishable, that ordering is now part of a group of $3!$ indistinguishable orderings. Paint the remaining blocks solid green and your have diminished the number of different orderings by another factor of $3!$.

Hence the answer for a) $\dfrac {6!}{3!3!} = \dbinom{6}{3,3}= \dbinom{6}{3}$

And by the same argument, starting with $9$ distinguishable blocks and gradually rendering groups indistinguishable, for b) $\dfrac {9!}{3!3!3!} = \dbinom{9}{3,3,3}$