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.


  • 1
    $\begingroup$ 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!. $\endgroup$
    – MrYouMath
    Jun 3, 2016 at 19:06

3 Answers 3


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


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) $\dfrac {9!}{3!3!3!} = \dbinom{9}{3,3,3}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.