Cfrgtkky

When we calculate the number of permutations of alike objects with the formula $$frac{n!}{p! cdot q! cdot r!}$$ when $p$ objects are of one kind, $q$ objects are of one kind, and $r$ objects are of one kind and so on, what are we actually getting?

We're getting the number of "distinct" permutations. An example might help: how many ways can we rearrange the letters in the word "mom"?

We can list them explicitly by brute-force:

• mom


• mmo


• omm


• mom


• mmo


• omm

... something seems odd about that list, no? I mean, why did I list the same things twice? Well, notice, if you swap any two of the m's that technically it's a different permutation of the letters, despite looking the same.

That is the motivating idea behind your formula. For any given word in the list above, we can rearrange the m's in it in $2!$ ways (which comes from the number of rearrangements of $2$ objects). The total number of permutations of a $3$-letter word (counting repeats as applicable) is $3!$. Thus: we divide by $2!$ to get the number of unique permutations:

$$text{total permutations} = 3! ;;; text{distinct permutations} = frac{3!}{2!} = 3$$

Indeed, notice, explicitly listing the number of permutations for "mom" and cutting out repeats, we indeed have $3$:

• mom


• mmo


• omm

Obviously, this idea generalizes easily to any number of letters (or whatever you're rearranging), and to having multiple groups of repeating things (we just add more factorials to the denominator). Doing this gives us the number of "distinct" or "unique" permutations.

If we set $5$ persons on a row then there are $5!=120$ possibilities for that.

If we have $5=r+g+b$ balls on a row from which $r=2$ are red, $g=1$ are green and $b=2$ are blue then at first hand there are $5!=120$ possibilities for that.

But if we can distinguish the balls only by color then we "see" less possibilities.

Results like $R_1G_1R_2B_1B_2$ and $R_2G_1R_1B_2B_1$ are look-alikes.

So if we are after distinguishable possibilities then we are overcounting, and result $RGRBB$ is counted exactly $2!1!2!$ times.

This shows that the overcounting can be repaired by dividing with factor $2!1!2!$ resulting in $frac{5!}{2!1!2!}=30$ possibilities.

Of course this can easily be generalized.

It is called Permutation of multisets.

Here is another way to look at it.

Let's say there are $r=3$ red, $g=4$ green and $b=2$ blue balls. How many ways can the balls be aligned in $n=3+4+2=9$ positions?

First, we can put the $3$ red balls in $9$ positions in ${9choose 3}$ ways. (Note that the order is not important as the red balls are alike (indistinguishable).

Second, we can put the $4$ green balls in the remaining $6$ positions in ${6choose 4}$ ways.

Third, we can put the $2$ blue balls in the last two positions in ${2choose 2}$ ways.

Hence, the total number of permutations of $9$ balls, of which $3,4,2$ are indistinguishable (alike), is:
$${9choose 3}{6choose 4}{2choose 2}=frac{9!}{3!cdot 6!}cdot frac{6!}{4!cdot 2!}cdot frac{2!}{2!cdot 0!}=frac{9!}{3!cdot 4!cdot 2!} text{or}\
{9choose 3,4,2}=frac{9!}{3!cdot 4!cdot 2!}.$$