# Counting in how many ways one can make a choice subject to some restrictions

I'm perplexed in front of the following questions. I hope I get clear explanation for it...

A box of biscuits contains 9 diff. varieties (e.g. A, B, C, D, E, F, G, H, I). In how many ways can 4 biscuits be chosen if: Two each of two varieties are selected?

The final answer is 36 but I don't how to get this answer as I don't understand exactly what does it mean "Two each of two varieties are selected" and I'd love to see several examples for it using the letters (no need to list all the 36 combinations of course - just a couple will be fine for me get the hang of the question). For example: is it something like (AB)(AB)? or (AB)(CD)? or what?

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The desired choice must result in biscuits of the form $$XXYY$$ where $X,Y \in \{A,B,C,D,E,F,G,H,I\}$ and $X \neq Y$.

Note that for us, $YYXX$ is the same as $XXYY$.

There are $9$ options to choose $X$.

Once this is done, $Y$ has only $8$ options.

Hence, the number of ways is $72$.

However, note that $XXYY$ is the same as $YYXX$.

Hence, we need to divide by $2$ to get the answer.

Hence, the answer is $\dfrac{72}2 = 36$.

All the possible options are $$\{A A B B,A A C C,A A D D,A A E E,A A F F,A A G G,A A H H,A A I I,B B C C,\\B B D D,B B E E,B B F F,B B G G,B B H H,B B I I,C C D D,C C E E,C C F F,\\C C G G,C C H H,C C I I,D D E E,D D F F,D D G G,D D H H,D D I I,E E F F,\\E E G G,E E H H,E E I I,F F G G,F F H H,F F I I,G G H H,G G I I,H H I I\}$$

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For example, I could choose two C and two F biscuits. Notice that this is considered the same as choosing two F biscuits and choosing two C biscuits.

It's a strange question because it's exactly the same as choosing just one biscuit of one type then another biscuit of a different type, but I suppose it might make more sense in context.

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