# Formula for combinations with repeting objects

1. Sep 12, 2011

### musicgold

Hi,

I am not sure which formula will allow me to get the following results.

If I have to pick two letters from A, B, C, D, then there are 6 possible combinations:
AB
AC
BC
BD
CD.

Now assume that letter D has turned into another A, i.e. A2. Now my combinations have changed to
A1B
A1C
A1A2
BC
BA2
CA2

So I eliminate BA2, and CA2 and I have 4 unique combinations.

I know that in the case of permutations with repetitions, we divide the nPr formula by the factorials of the frequency of the letters, but I am not sure how it is done with combinations. What formula can use for combinations?

Thanks.

2. Sep 12, 2011

### HallsofIvy

Staff Emeritus
Yes, that is
$$\left(\begin{array}{c} 4 \\ 2\end{array}\right)= \frac{4!}{2! 2!}= 6$$

Now assume that letter D has turned into another A, i.e. A2. Now my combinations have changed to
A1B
A1C
A1A2
BC
BA2
CA2

So I eliminate BA2, and CA2 and I have 4 unique combinations. [/quote]
Okay, with the "1" appended you still have 4 objects but without that, you cannot distinguish between combinations where the "A"s have been swapped. However, here "A1A2" and "A2A1" which are two distinct permutations are NOT two distinct combinations.
If you were counting permutations, then it would be 6/2= 3 but with that "A1A2" and "A2A1" put back in there are 4 combinations.

3. Sep 12, 2011

### musicgold

Thanks HallsofIvy.

Your answer is not clear to me. As I said, I know the formula for the combination 4C2.
I am interested in the formula for the case where two letters are the same.