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

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

My try:

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.

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.

• Sir one doubt at last you are doing 2^2 , but shouldnt it be just 1*1 as such shoes are like xx ,yy in the incomplete ones we just have one way of choosing a incomplete one from them isnt just xy isnt? Apr 8, 2022 at 22:19
• @Buraian check this out Apr 8, 2022 at 23:54
• @ProblemDestroyer I don't quite understand what you want to say, things like $xx$, $xy$ have no meaning intrinsically so you need to explain what you mean to say by this. What is important here is that each shoe is distinguished (it both belongs to a specific pair and within the pair it is either the left or the right shoe), is a choice is really that of a subset, of size $6$, of the set of $16$ shoes (though not all choices are allowed). That means for each pair from which $1$ shoe is taken, we must know which one it is, and that gives $2$ distinct possibilities for each such pair (cont'ed..) Apr 9, 2022 at 9:29
• ... Maybe you think a choice is determined by just telling for every pair whether $0$, $1$ or $2$ of the shoes are chosen, but that is not the case (if it were socks, where we often cannot tell left from right, that might be a plausible interpretation but the question should still make this clear; for shoes it is excluded). If that had been the problem, it would amount to counting the distinct permutations of the sequence $[2,2,1,1,0,0,0,0]$, which is $420$, but each of these gives $\binom22^2\binom21^2\binom20^4=4$ actual solutions. Apr 9, 2022 at 9:37

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 $28\cdot60=1680$.