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.


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.


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.

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

  • $\begingroup$ 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: $\frac{\binom{10}{1}\binom{9}{6}2^68!}{}$. 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.