Formula for combinations with repeting objects

[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
AD
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.

 

[HallsofIvy]

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
AD
BC
BD
CD.

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.

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.

 

[musicgold]

Thanks HallsofIvy.

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.

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.