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While choosing 2 from 3, we do ${3 \choose 2}$. But what would happen if we do ${3 \choose 1}{2 \choose 1}$? [Choosing 1 from 3 and again 1 from the remaining two)?

I know the latter is incorrect but can anyone give me conceptual view of how that's wrong and what would I be doing if I did the latter?

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${3 \choose 2}$ chooses 2 objects out of three without paying attention to the order in which they were chosen. It only matters which objects are chosen.

${3 \choose 1}{2 \choose 1}$ chooses 2 objects sequentially, so the order does matter.

Example: the three objects are ABC.

The first method counts the number of sets of two that can be chosen: $\{A,B\},\{A,C\},\{B,C\}$.

The second method counts the number of 2 letter words that can be built from different letters: AB, BA, AC, CA, BC, CB.

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You are actually overcounting on the second one. Consider three objects $A$, $B$, and $C$. If you choose one from the three objects and then choose another one from the remaining two, you could actually pick the same two items in two different ways. For example, you could pick $A$ first, then from the remaining two objects($B$ and $C$) you pick $B$. But you could also pick $B$ first then $A$. But if you were choosing 2 objects at the same time from the 3 objects, then it doesn't matter if you pick $A$ or $B$ first, you technically have the same thing.

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C(3,2) is combinations, so order doesn't matter. C(3,1)C(2,1) is the number of permutations. This latter one is because you are saying that first you need to pick one, and then count how many ways you can pick from the others. So, picking from abc would sample from ab, ac, ba, bc, ca, and cb. For C(3,2), you remove the reorderings, so ab and ba are identical, and same with bc/cb and ac/ca. This leaves us witb only 3 options, not 6.

You are actually approaching how to count permutations. Instead of C(n,k), we start with C(n, 1) to pick the first location, and repeat this. So for permutations P(n,k), it is the same as $C(n,1)*...*C(n-k,1) = n*...*(n-k)= \frac{n!}{k!}$

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