Why do we divide values to count permutations with repeated letters? My question is why we divide values in permutations in case of repetition of words
Like if we have to arrange the word "coffee" in different ways then my answer will be 6!/2!x2! (i know the complete formula). i cant understand why we divide in case of repetitions of word. please clarify.
Thanks
 A: The purpose of dividing is to reverse a multiplication which has taken place, which should not have taken place.
Generally, the multiplication "should not" have taken place because two outcomes are not "distinct", i.e. they are the same.
For example if I ask how many distinct ways two coins can turn out, you will say there are 2 ways for the first coin, and 2 ways for the second so the answer is $2\times2=4$.
However if I then tell you that one of the coins is a double-headed coin, you now know that the 2nd multiplication by 2 is in error, as there is only one distinct way that coin can turn out.  So you divide by $2$ to give your answer $\frac{4}{2}=2$.
In this example it's obvious the answer is two but sometimes when you write out a solution it's easier to multiply all the possibilities first and then divide to "uncount" the ones that aren't distinct.
A: Assume for the moment you have the six letters $c$, $o$, $f_1$, $f_2$, $e_1$, $e_2$ to arrange. This can be done in $6!$ ways. After arrangement of these six letters a first assistant appears who erases the indices of the $f$s, making every one of these $6!$ arrangements indistinguishable from exactly one other of these arrangements. It follows that now there are ${6!\over2}$ indistinguishable arrangements left over. Now a second assistant appears who erases the indices of the $e$s, making every one of the momentaneous arrangements indistinguishable from exactly one other of these arrangements. After this second pruning there are ${6!\over2\cdot2}$ indistinguishable arrangements left over.
A: A tree showing permutations for a set of {n} = 5 and selecting subset of {n} = 3
Adding to the above answers, I would like to add in the intuition for as to why we divide by the number of ways of getting the permutations which are same to us, in our case due to repeating letters.
I will start with numbers and then we will talk about "coffee".
In the image you can see the set of permutations, imagine it to be a bag full of things of our interests. These "things" can be divided into groups. For example, if for some reason, to us there is no difference if subsets start with 1, 2, 3, 4 or 5 as far as combinations are concerned, then the total combinations relevant to us will be $\frac{^5P_3}{3}$, that is, $\frac{1}{3}$rd of the total permutations. In other words there are 3 ways to get permutations which either start with 1, 2, 3, 4 or 5, which are, place one of the remaining 3 numbers at the first spot after you fill the last two spots. This means we have 3 groups which are identical to each other, each having 20 members.
(Refer to second image for better understanding)
3 identical groups of 5P3/3
Going further. For a set $Z$ with $ n\{Z\}=N$,  where all numbers are unique, if we want all the subsets $X$ with $n\{X\}=k$, meaning that ordering of numbers in subsets is not important for us, we'll have $$p = \text{ } ^NP_k \text{} \text{   ,where } p \text{ is the Total permutations }$$
Now these permutations $p$ will have $k!$ groups, each group representing a way of permutation. Each group contains our desired combinations ordered in the particular way which the given group represents. The differentiating factor between the members of a group with members of another group is the order/permutation which makes no difference for us, meaning $$\frac{p}{k!}=c \text{} \text{ , where } c \text{ is the Total combinations}$$
This is nothing but $$C(n,k)=\frac{N!}{(N-k)!\times(k!)}=\frac{P(n,k)}{k!}$$
Now same logic applies for "coffee". The exchanging of places of 'e' and 'f' doesn't matter to us, they are the same letter. So number of ways to switch one 'e' with the other 'e' is $2!$ i.e. $2!$ identical groups form. Now after selecting one of the groups we have selected a group which:
a) Now correctly doesn't distinguishes between the two 'e's and
b) Considers the 2 'f's to be different and their relative order in the word to be important!
This is solved by again dividing by the number of ways two 'f's exchange their positions i.e. $2!$. Ultimately giving us, $$\frac{6!}{2!\times2!}$$
