Permutations and Combinations - and repeating letters case

Question

I am having a hard time understanding a simple concept - of arranging alphabets in a word where there are repetitions. Specifically, the part where we divide the number of permutations by the factorial of the repetition.

From the textbook :

In general, repetitions are taken care of by dividing the permutation by the factorial of the number of objects that are identical.

If you look at the word TOOTH, there are 2 O’s in the word. Both O’s are identical, and it does not matter in which order we write these 2 O’s, since they are the same. In other words, if we exchange 'O' for 'O', we still spell TOOTH. The same is true for the T’s, since there are 2 T’s in the word TOOTH as well. In how many ways can we arrange the letters in the word TOOTH?

We must account for the fact that these 2 O’s are identical and that the 2 T’s are identical. We do this using the formula:

nPr/x1!x2!, where x is the number of times a letter is repeated.

My question is why ? Or rather how ?

Why do we divide the total number of permutations by the FACTORIAL of the instances of repeating elements ? I am missing a link in the thought process, so any insight is welcome.

Answer 1

In a field of sheep a shepherd has difficulty distinguishing and counting the heads and bodies of the sheep from one another... (they are too fluffy) but can clearly see all legs of the sheep and count them. He knows that every sheep has four legs and so when counting the legs he counted a total of $40$ legs for the sheep. He knows then that there were exactly $\frac{1}{4}\times 40 = 10$ sheep in his flock.

The "shepherd's principle" is a counting principle which says that if we want to count the number of objects in a scenario and we counted $x$ objects but we had overcounted each object multiple times for a total of exactly $y$ times for each object... then we may correct our count by dividing the total number we had initially counted with repeats by the number of times each object was repeated in our count for a corrected total of $\frac{x}{y}$ distinct objects.

Here, if we were to temporarily assume each letter was distinct... like $T_1O_1O_2T_2H$... and arrange those, we will notice that we had overcounted the outcomes. Specifically, for each relabeling of repeated letters, we will have counted that despite the unlabeled arrangements being otherwise identical. We divide by the amount we overcounted by, namely $2!\cdot 2!$.

Another explanation... we first choose the positions that the $T$'s occupy simultaneously, followed by the positions the $O$'s occupied, etc... for a total of $\binom{5}{2}\times\binom{3}{2}\times\binom{1}{1}$ which is equal to $\frac{5!}{2!2!1!}$, and in general with non-negative integers $n_1,n_2,\dots ,n_k$ such that $n_1+n_2+\dots+n_k = N$ we have that $\binom{N}{n_1}\binom{N-n_1}{n_2}\binom{N-n_1-n_2}{n_3}\cdots\binom{N-n_1-n_2-\dots-n_{k-1}}{n_k} = \dfrac{N!}{n_1!n_2!n_3!\cdots n_k!}$