**Suppose we have to find the
no. of permutations by arranging the four letters of the word DOOR. **

If we apply simple permutation formula here, our answer will be 4!

But this is wrong

Why?

It is because O comes two times in every word which is made by shuffling these letters.

For example, ORDO

Now, the problem is that we are considering two Os as different
letters, say O_{1} and O_{2}.

Then, O_{1}RDO_{2} and O_{2}RDO_{1}
will be counted differently, but actually these two are same!!

So, we must have to remove these extra counts. Now each word can be written in two ways (if we consider the two Os differently), so we have to divide 4! by 2! to get the actual answer.

Let us understand it in more detail.

Suppose we have a word **reese**.

We have three e’s in this word.

Suppose we have to make a three letter word from the given word. Then in how many ways we can do so?

Normaly, if we consider these e’s as different letters, then we have 5×4×3 ways of doing the task.

But, we have 6 ways to write each word (if we are considering the e’s differently)

Example,

**So we have to count 6 words as 1**

**So we have to divide our answer by 6**

In general, if we have any thing which is being repeated in a word or permutation, then we have to divide the obtained answer by factorial of n. Where n is the no. of times that thing is repeated!!

What if two or more things are being repeated?

//please do not get confused reading the paragraph given below,

Just chill and read the next lines//

Then we will obtain our answer for first thing by the above formula, and then after doing it we are left with the no. of permutations, considering two or more permutations ( where the first thing is being treated distinctly) as one. Now while counting these permutations, the other things (which are also repeating) are considered distinctly, so we have to again repeat the above step.

##So confusing##

No, it’s actually not

In short,

The number of permutations

of n objects of which p1

are of one kind, p2

are of second kind, …, pk

are of kth kind and

the rest if any, are of different kinds is

Now let’s do a question by using the above method

Q Find the number of permutations of the letters of the word ALLAHABAD.

Ans we have total 9 places

First place can be filled by any of the letters, second by the remaining and so on.

So, total permutations are 9!

Hey stop stop stop !!!

Why?

Did we checked if any letter is repeating?

No

So check

Ok, in the letter ALLAHABAD, we have 4 A’s, 2 L’s.

So what we are supposed to do?

Yes, you are right, we must divide 9! by 4! and 2! to get the actual answer.

So our answer will be