# You have letters: $A,A,C,D,E$ in a bag. You pick $3$ at random without putting them back. What is the total number of combinations you can make?

You have letters: $$A,A,C,D,E$$ in a bag. You pick $$3$$ at random without putting them back. What is the total number of combinations you can make?

I am doing $$\binom{5}{3}$$, however, this results in $$10$$. I know this must be incorrect because these are the list of all combinations possible: $$[a,a,c],[a,a,d],[a,a,e],[a,c,d],[a,c,e],[a,d,e],[c,d,e]$$

I would really appreciate someone showing me how this kind of problem is solved. It is basic but I can't seem to wrap my head around what's happening.

• The answer is not ${5 \choose 3}$ because the letter $A$ appears two times. You can solve by splitting into two cases, in the first case consider all combinations with only one $A$ and in the second case consider all combinations with two $A$'s. Sep 17 at 10:02
• Sure, I understand that. This does indeed get the right answer. However, I was unsure if it's possible to go straight to the final value. Sep 17 at 10:04
• @Asher2211 Minor quibble: "...all combinations with at most only one A...". Sep 17 at 10:17
• See my answer for a "... straight ... final value". Sep 17 at 10:20

Two cases arise:

Case I: Both the $$A$$ are selected. In this case, you will choose the remaining (third letter) in $$\binom{3}{1}=3$$ ways.

Case II: In this case, you selected $$3$$ letters out of the $$4$$ letters, $$A,C,D,E$$. You will do this in $$\binom{4}{3}=4$$ ways.

So, there are $$3+4=7$$ ways.

Alternative approach:

The only over-counting is when there are exactly $$2$$ A's. In this case, the over-counting is exactly double.

Therefore, the shortcut is $$\displaystyle \binom{5}{3} - \binom{3}{1}.$$

• Yes, that's perfect! I understand that now. I would upvote your answer because I personally prefer it more but I don't have enough reputation but thank you. What exactly does the 3C1 represent above? Sep 17 at 10:24
• Let me upvote him on your behalf. Sep 17 at 10:25
• @SyedArafatQureshi How many ways are there to form a combination that has exactly $2$ A's. Answer : $\displaystyle \binom{3}{1}.$ Therefore, within the $\displaystyle \binom{5}{3}$ enumeration, exactly $\displaystyle \binom{3}{1}$ combinations have inappropriately been counted twice. Sep 17 at 10:28
• @user2661923 Thank you. It makes sense now. Sep 17 at 10:30

Another way of looking at this, that I think separates the cases clearly, is:

Three cases arise:

1. no $$A$$ is selected which gives $$\binom{3}{3}=1$$ selection (all three of $$C,D,E$$)

2. one $$A$$ is selected, which gives $$\binom{3}{2}=3$$ selections (two of $$C,D,E$$)

3. both $$A$$s are selected, which gives $$\binom{3}{1}=3$$ selections (one of $$C,D,E$$)

This gives $$1+3+3=7$$ possibilities.