# shoes probability for finding exact k pairs of shoes

The closet has $10$ pairs of shoes. The no. of ways in which $6$ shoes can be choosen from it show that There will be

(i) no Complete pair

(ii) exactly one complete pair

(iii) 2 complete pair

(iv) 3 complete pair

My Try:

Total no. of selecting $6$ shoes from $20$ shoes is $\displaystyle = \binom{20}{6}$

Now for first part...

First we will select $6$ pairs from $10$ pairs is $\displaystyle = \binom{10}{6}$

Like $6$ pairs is $(A_{1},A_{2})\;\;,(B_{1},B_{2})\;\;,(C_{1},C_{2})\;,(D_{1},D_{2})\;,(E_{1},E_{2})\;,(F_{1},F_{2})$

Noew select $1$ shoes from each pair is $= \displaystyle \binom{2}{1}\times \binom{2}{1} \times \binom{2}{1} \times \binom{2}{1}\times \binom{2}{1}\times \binom{2}{1}=2^6$

So Required probability for first is $\displaystyle \frac{\binom{10}{6}.2^6}{\binom{20}{6}}$

Now can anyone explain me how can i solve other parts

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HINT: To get exactly one complete pair, select a pair, and then select $4$ non-matching shoes from the remaining $9$ pairs. There are $10$ ways to choose the pair, and the rest of the computation is like that of (i), which you already know how to do.