Question 1025274
<pre>
Take the word DEED

If we make one D capital and the other small d and
do the same with the E, we have these 
4! = 4*3*2*1 = 4P4 = 24 permutations:

1.  DdEe
2.  DdeE
3.  DEde
4.  DEed
5.  DedE
6.  DeEd
7.  dDEe
8.  dDeE
9.  dEDe
10.  dEeD
11.  deDE
12.  deED
13.  EDde
14.  EDed
15.  EdDe
16.  EdeD
17.  EeDd
18.  EedD
19.  eDdE
20.  eDEd
21.  edDE
22.  edED
23.  eEDd
24.  eEdD

For instance, we can tell the difference between 

EDed, EdeD, eDEd, and edED

But if we spell them all with capital letters we cannot
tell them apart.  They all look like EDED.  So we cannot
distinguish them, so we label them "indistinguishable".
However we can tell the difference between EDED and DEDE.
They are "distinguishable".

So there are only 6 distinguishable permutations of DEED.

They are

1. DDEE
2. DEDE
3. DEED
4. EDDE
5. EDED
6. EEDD

That's because the 4! = 4*3*2*1 4P4 = 24 counts each one
2! = 2*1 = 2P2 = 2 times too many for the D's and also
2 times too many for the E's.  Tham means that the 24
counts each permutation 4 times too many.  So to get the 
6, we begin with the 24! and divide by the product of each 
of the factorials of the numbers of indistinguishable letters.
So we divide 24 by 2! and again by 2! so we end up dividing
24 by 4 getting 6.

{{{4!/(2!2!)}}} = {{{24/(2*2)}}} = {{{24/4}}} = {{{6}}}

The number of distinguishable permutations of the word PEPPER
is {{{6!/(3!2!)}}} = {{{720/(6*2)}}} = {{{720/12}}} = {{{60}}}

We divide 6! by 3! for the indistinguishable P's and by 2!
for the indistinguishable E's.

The number of distinguishable permutations of the word MISSISSIPPI
is {{{11!/(4!4!2!)}}} = {{{34650}}}

We divide 11! by 4! for the indistinguishable I's, by 4! for the
for the indistinguishable S's, and by 2! for the indistinguishable
P's.

Edwin</pre>