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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.

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  • $\begingroup$ 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. $\endgroup$
    – Asher2211
    Sep 17 at 10:02
  • $\begingroup$ 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. $\endgroup$ Sep 17 at 10:04
  • $\begingroup$ @Asher2211 Minor quibble: "...all combinations with at most only one A...". $\endgroup$ Sep 17 at 10:17
  • $\begingroup$ See my answer for a "... straight ... final value". $\endgroup$ Sep 17 at 10:20
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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.

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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 $\displaystyle \binom{5}{3} - \binom{3}{1}.$

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    $\begingroup$ 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? $\endgroup$ Sep 17 at 10:24
  • $\begingroup$ Let me upvote him on your behalf. $\endgroup$
    – Ashiq
    Sep 17 at 10:25
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    $\begingroup$ @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. $\endgroup$ Sep 17 at 10:28
  • $\begingroup$ @user2661923 Thank you. It makes sense now. $\endgroup$ Sep 17 at 10:30
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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.

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