The number of ways of selecting 6 shoes from 8 pairs of shoes so that exactly 2 pairs of shoes are formed?

My try:

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Let us first choose $2$ pairs from $8$ pairs. It can be done in $_{8}C_2$ ways.

Suppose I have chosen pair $(1,2)$ and $(3,4)$. We have to chose other 2 shoes such that one is not a pair of another.

For example if I choose 5 I shouldn't choose it's pair ie 6. It can be chosen in $_{12}C_1 \cdot _{10}C_1$ ways. So total no. of ways = $_8C_2\cdot _{12}C_1\cdot_{10}C_1=3360$ ways.

Am I correct. Please clarify.


No you are not correct in your final steps. Here is how you could proceed instead: first choose two complete pairs out of $8$ (as you did), for $\binom82=28$ choices. Then choose two pairs from the remaining $6$ for which you will (later) make two incomplete pairs, for $\binom62=15$ choices. Finally from each of those two incomplete pairs, choose one shoe, for $2^2=4$ choices. Every final outcome can be obtained in exactly one way (this is where it differs from your method) and one gets $28\times15\times4=1680$ possibilities.

You found twice that. That is because you found each solution in two different ways, for the $2!=2$ different orders in which you can choose the two unpaired shoes.


When unsure, try with a smaller instance of the problem.

Let's count the ways of choosing 2 shoes from 2 pairs so that they are not in the same pair. Your method would say start with 0 full pairs out of 2 and then multiply by $_4C_1$ and $_2C_1$:

$$\binom{2}{0} \binom{4}{1} \binom{2}{1} = 8$$

But we can count directly: out of $ABab$, the choices fulfilling our conditions are $Aa$, $Ab$, $Ba$, $Bb$. That's not 8 so the formula is not correct.

At the same time it shows what should have been done differently and why: divide your result by 2 to reflect that the last 2 shoes that must not be from the same pair can be picked in any order ($aA = Aa$).


As you said you can choose $2$ pairs out of $8$ pairs in ${8\choose2}=28$ ways. When these $2$ pairs have been chosen there are $6$ complete pairs left, and you have to choose another $2$ shoes out of these such that no additional pair is formed. This is possible in ${12\choose2}-6=60$ ways. The total number of admisible choices therefore is $28\cdot60=1680$.


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