1
$\begingroup$

The question is:

How many ways can you align 3 red balls, 2 blue balls and 2 yellow balls in a row given that at the beginning and the end of the row there is a red ball? (balls of the same color aren't unique).

My solution was - since we have to put two red balls at the beginning and end then we need to order only 1 red ball, 2 blue and 2 yellow. Let's say the order matters, then there are 5! = 120 ways to order the balls. Since we counted as though 2 blue balls differ we need to correct that and divide by 2. Same goes for the yellow balls.

Then the answer is 30.

My professor came up with a different way when he published the answer. Simply: (5 choose 2,2,1) = 30. He didn't explain what this is and how to compute it.

I'm assuming it's an abbreviation for saying - from 5 balls choose 2 blue, 2 yellow and 1 red but can't find any reference to it. Can someone help?

$\endgroup$
5
$\begingroup$

See multinomial coefficient for an explanation:

It is denoted, $$\displaystyle \binom{n}{k_1, k_2..., k_m} = \dfrac{n!}{k_1! \times k_2! \times \cdots \times k_m!}$$ and in your case, $(5\;\text{choose}\, 2, 2, 1)$ is denoted and computed as: $$\binom{5}{2, 2, 1} = \frac{5!}{2!\,2!\,1!} = \dfrac{120}{2\cdot 2 \cdot 1} = \dfrac{120}{4} = 30$$

Recall, for any number $n,\,$ $n\,! = n(n-1)(n-2)\cdots(2)(1)$.

$\endgroup$
2
$\begingroup$

After placing a red ball in front and a red ball behind, one can place the remaining two blue, two yellow and one red ball in $\frac{5!}{2!2!1!}$ ways. This is an example of a multinomial coefficient.

$\endgroup$
1
$\begingroup$

The multinomial coefficient your professor is referring to is interpreted as follows:

$${n\choose{k_1, k_2, ..., k_m}}=\frac{n!}{k_1!\times k_2!\times...\times k_m!}$$

$\endgroup$

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.