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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.

The same idea will work for the other two problems, except that in (iv) it’s even easier.

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