Permutations when there are pairs or similar units

Question

There are four colours (red, red, blue, green) assigned to a group of four people. What are the number of possible combinations?

I know the answer is 12 but I would really appreciate if someone explained it using a simple formula. I know that 4x3x2x1=24, then dividing that by two (since the two reds can reside in each other's positions) would give 12. However, is there a formula to solve more complicated sets that have many pairs of similar colours.

What would be the number of combinations in a set of 10 consisting of five red, four blue, and one white?

Answer 1

What you want is permutations from a multiset. For example: $$ A=\{n_1\cdot a_1, n_2\cdot a_2, \cdots , n_n a_n \} $$ Is a set with $n_1$ repetitions of the element $a_1$, $n_2$ repetitions of element $a_2$, and so on...

When you want to count all possible permutations of the set $A$, you need to do what you've concluded in your post. You suppose all elements are different and permute them all $(n_1+n_2+\cdots + n_n)!$. After that, you consider that there are in fact repeated elements and remove the permutations between those repeated elements:$(n_1)!(n_2)!\cdots (n_n)!$. Therefore your final answer is: $$ {n_1+n_2+\cdots+n_n \choose n_1,n_2,\cdots ,n_n} = \frac{\big(\sum_{i=1}^{n} n_i\big)!}{\prod_{i=1}^{n}(n_i)!} $$ So answering your last question: $$ {5+4+1 \choose 5,4,1} = \frac{(10)!}{5!4!1!} = \frac{3628800}{2880} = 1260 $$

Answer 2

It would be

$$\frac{10!}{5!\cdot4!}$$

Same reasoning, really: if all the $10$ objects were different, it would be $10!$, but since the $5$ reds are the same, you need to divide by $5!$, and similarly for the $4$ blue ones.

Answer 3

It works the same way with repeated elements. Suppose you wanted to know all the permutations of the word MISSISSIPPI. You have I repeated 4 times, S repeated 4 times, and P repeated twice. The total number of permutations of the letters of MISSISSIPPI is $11!$. But since there are $4$ I's and S's, and $2$ P's, you divide $11!$ by all of $4!,4!,$ and $2!$.

$$\frac{11!}{4!4!2!1!}$$

For your question, you have $10$ objects, $5$ reds, $4$ blues, and $1$ white. Thus

$$\frac{10!}{5!4!1!}$$