# Thread: Permutations

1. ## Permutations

Theres one kind of problems dealing with permutations. I know the solving method of these kind of problems but I dont get why does it work.
Heres an example:
Lets say we are given a word SCIENCE and we want find in how many ways can we rearrange letters of the word (how many permutations are there).
So, therere 7 letters in the word. If each letter was unique number of permutations would be 7! But theres 2 C and 2 E. So we need to divide 7! by number of permutations which these letters will make. We exclude repeated permutations (e.g. swapping places of first C and second doesnt change anything).
So the formula will look like: 7!/(2! * 2!)
I dont understand why do we divide. It doesnt make any sense for me. I understand that we need to exclude repeating permutations but why does dividing work? I cant grasp an intuition behind it

P.S. Im not a native English speaker and Im sorry if I couldnt explain what I wanted well enough.

2. ## Re: Permutations

Consider a smaller word like POT. There are 3! = 6 permutations:

P0T PTO TOP TPO 0PT 0TP. Now what if the T had been a P.

Try literally replacing the T's by P's. You get:

POP PPO POP PPO OPP OPP.

There are half as many distinct ones. Why? Because, for example POT and TOP which were different are now both POP, which are the same.

3. ## Re: Permutations Originally Posted by Jotaro Theres one kind of problems dealing with permutations. I know the solving method of these kind of problems but I dont get why does it work.
Heres an example:
Lets say we are given a word SCIENCE and we want find in how many ways can we rearrange letters of the word (how many permutations are there).
So, therere 7 letters in the word. If each letter was unique number of permutations would be 7! But theres 2 C and 2 E. So we need to divide 7! by number of permutations which these letters will make. We exclude repeated permutations (e.g. swapping places of first C and second doesnt change anything).
So the formula will look like: 7!/(2! * 2!)
I dont understand why do we divide. It doesnt make any sense for me. I understand that we need to exclude repeating permutations but why does dividing work? I cant grasp an intuition behind it
Consider the phrase $LOOKGOOD$. If we add subscripts to the repeated letters $LO_1O_2KGO_3O_4D$
we now have a string of eight distinct letters that can be rearranged in $8!$ ways.
Now the four subscriped letters $O_1 O_2 O_3 O_4$ can themselves be rearranged in $4!=24$ ways.
To account for the twenty-four duplications without subscripts we divide.
Thus without subscripts $LOOKGOOD$ can be rearranged in $\dfrac{8!}{4!}$ ways.

4. ## Re: Permutations Originally Posted by Walagaster Consider a smaller word like POT. There are 3! = 6 permutations.
...or consider 123: still 3!, right?