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I came across following problem from Sheldon Ross' book:

A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the probability that there will be

  1. no complete pair;
  2. exactly one complete pair

I solved them as follows:

Problem 1

  • 1st shoe can be anyone out of 20: $\frac{20}{20}$
  • 2nd shoe can be anyone out of remaining 19 except the one forming pair with previously selected shoe: $\frac{18}{19}$
  • 3rd shoe can be anyone out of remaining 18 except those forming pair with previously selected shoes: $\frac{16}{18}$

and so on till we choose 8 shoes. We need to take multiplication of all, leading to $\frac{20\times18\times16\times14\times12\times10\times8\times6}{20\times19\times18\times17\times16\times15\times14\times13}=0.091$

Problem 2

Following same logic of problem 1,

  • 1st shoe can be any one out of 20: $\frac{20}{20}$
  • 2nd show should be the one forming pair with 1st one
  • 3rd shoe can be any one out of remaining 18: $\frac{18}{18}$
  • 4th show can be any one out of remaining 17 except the one forming pair with previously selected one: $\frac{16}{17}$

and so on till we choose 8 shoes. We need to take multiplication of all, leading to $\frac{20\times1\times18\times16\times14\times12\times10\times8}{20\times19\times18\times17\times16\times15\times14\times13}=0.015$

It turns out that the solution to first problem is correct, but the solution to second problem is incorrect. Its given as follows: $\frac{\binom{10}{1}\binom{9}{6}\color{red}{\frac{8!}{2!}}2^6}{20\times19\times18\times17\times16\times15\times14\times13}$

I understand we can select one pair out of 10 in $\binom{10}{1}$ ways. We select both shoes from this pair. Then we can select six pairs out of remaining nine in $\binom{9}{6}$ ways. We can select any one out of two shoes of each of six pairs in $2^6$ ways. But I dont understand from where $\frac{8!}{2!}$ came.

Edit

The solution given at the back of the book is $0.4268$. But books solution manual solves it as $\frac{\binom{10}{1}\binom{9}{6}\color{red}{\frac{8!}{2!}}2^6}{20\times19\times18\times17\times16\times15\times14\times13}$ which I just checked to be equal to $0.2133$. This pdf gives the solution as $\frac{\binom{10}{1}\binom{9}{6}\times 2^6}{\binom{20}{8}}$ which matches with $0.4268$. So now I am guessing what is correct answer and how can I get the answer for problem 2 by following same approach as I followed for problem 1.

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Your text is in error.     There should be no $8!/2!$ factor.

In general the probability for selecting $x$ pairs $(x\in \{0,1,2,3,4\})$ is$$\dfrac{\dbinom{10}{x}\dbinom{10-x}{8-2x}2^{8-2x}}{\dbinom{20}8}\quad\text{or}\quad\dfrac{\dbinom{10}{8-x}\dbinom{8-x}{x}2^{8-2x}}{\dbinom{20}8} $$

The series sums to 1 as required.


Thus the probability for selecting no pairs is $\binom{10}{8}2^8/\binom{20}{8}$, which is equal to your answer.

And the correct probability for selecting exactly one pair is $\binom {10}{1}\binom{10-1}{8-2}2^{8-2}/\binom{20}{8}$.   Alternatively that is $\binom{10}{8-1}\binom{8-1}12^{8-2}/\binom{20}8$.


By your method you can select six from the 10 pairs, one shoe from each of those, and both shoes of one from the remaining pair.

$$\dfrac{20\cdot18\cdot 16\cdot14\cdot12\cdot 10\cdot 4/6!}{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13}$$

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  • $\begingroup$ yess that is correct answer, can you please have a look at edit I added at the end of my question? I want to know whether it is possible to solve 2nd problem with the same approach as I followed for 1st problem, that is multiplication of probabilities of successively selecting shoes instead of using binomials. $\endgroup$ – anir Jun 16 '18 at 14:08
  • $\begingroup$ ohkay it seems that it should be $8!$ instead of $\frac{8!}{2!}$, right? $\endgroup$ – anir Jun 16 '18 at 14:10
  • $\begingroup$ Thanks for the new approach you stated at the starting. I have two doubts: (1) Is that $\frac{8!}{2!}$ absolutely senseless? I mean replacing it with $8!$ gives same final result. But, I don't feel its logically correct. The solution looks like this: $\frac{\binom{10}{1}\binom{9}{6}2^68!}{20.19.18.17.16.15.14.13}$. Here the numerator follows pattern of solution $\frac{\binom{10}{1}\binom{9}{6}\times 2^6}{\binom{20}{8}}$ stated in the edit added to the question, while denominator is following [continued...] $\endgroup$ – anir Jun 16 '18 at 16:13
  • $\begingroup$ [...continued] the pattern of solution $\frac{20\times18\times16\times14\times12\times10\times8\times6}{20\times19\times18\times17\times16\times15\times14\times13}$, I gave for problem 1. (2) How $\frac{4/6!}{14.13}$ = "both shoes of one from remaining pairs" $\endgroup$ – anir Jun 16 '18 at 16:13
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This is a problem similar to deck cards and hands problems.

The twenty shoes may be seen as a deck of two color (Left and Right) and 10-suits.

A "hand" contains 8 cards. It may contain 0 pairs, 1 pair... to 4 pairs, and we are asked to evaluate the chances to get some of these types.

the total number of hands is $ \binom {20}{8}$

i) for no pairs, we have $\binom {10}{8} $ for the choice of the 8 distinct values and $2^8$ for the eight color choices (left or right). The probability of such a "hand" is then 0.0914

ii) for one pair, we have to choose it among ten possibles. Then we have to chose another 6 singles having distinct values from 9 values that remained. Then we have 2^6 choices for the color, for a total $\binom {10} {1} \binom {9}{6} 2^6$. The probability to get a hand like this is 0.4267

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