Question

If I have a set of letters eg ABBCCDEEEF

How many unique combinations of letters are there if I choose n letters

Answer 1

Is my question really hard or something XD

Answer 2

Ok then what about if I choose 9 out of 10 letters

Answer 3

ABBCCDEEEF
We have 10 letters. We need to find no. of unique combinations if 9 is chosen from 10
Let's say A is not chosen, since there are 2B, 2C, 3E---> we need to find the n. of permutations by 2!2!3!
then no. of unique combinations= \(\frac{9!}{2!2!3!}\)
if one B is not chosen=\(\frac{9!}{2!3!}\)
if one C is not chosen=\(\frac{9!}{2!3!}\)
if D is not chosen= \(\frac{9!}{2!2!3!}\)
if one E is not chosen=\(\frac{9!}{2!2!2!}\)
if F is not chosen=\(\frac{9!}{2!2!3!}\)
Total no. of combinations= sum of these

Answer 4

Sorry
*Let's say A is not chosen, since there are 2B, 2C, 3E---> we need to divide the no. of permutations by 2!2!3!

Answer 5

Okay thanks, the problem is I need to do this for choosing 8, 7, 6 ... 1 :D

Answer 6

Is the solution gonna be messy

Answer 7

Total no. of combinations = (no of characters)! /Product of factorial of no. of each repetition

no. of characters=10
repetitions:
B-2
C-2
E-3


so no. of possible arrangements = N

\[N = \frac{10!}{2!2!3!} \]

Answer 8

@apoorvk, is this the number of combinations if we choose 9?

Answer 9

"9". I don't get it? choose 9 what?

Answer 10

choose 9 letters out of the set of 10.

I'm trying to get how many unique ways there are of choosing 9 letters out of the set of 10 with repetitions

Answer 11

okay then... you need 10C9. for choosing 9 outta ten. and then the same denominator (for repititions)

Answer 12

OK, I'll give you the full problem. Using the fact that every number can only be broken down into prime factors one way, I'm trying to calculate the number of factors of a large number such as 54504450, which is equal to 2x3x3x5x5x7x11x11x11x13. Now if we can get the number of unique combinations of choosing 10, 9, 8...1 of these factors, then the total should give us the number of factors. I think

Answer 13

I think @ash2326 has got onto something here