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This is not HW just a practice question from the text.

Q.. (NOTE: I find that both part a and b say the same thing but the answers are different) Ten women attend a business luncheon. Each women checks her coast and a attache case. Upon leaving, each women is given a coast and case at random.

a. In how many ways can the coats and cases be distributed so that no woman gets EITHER of her possessions.

ans.$(d_{10})^2$ Would like to know why is this squared

b In how many ways can they be distributed so that no woman gets back both of her possessions.

ans. $n! * S(m,n)$ where $m= 2, n=10$ this ends up being equal to $0$ as $m<n$ (where $S(m,n)$ is stirling number of the second kind)

why are these two questions different and why is part a $(d_{10})^2$

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1 Answer 1

up vote 2 down vote accepted

Let $\sigma$ be a permutation of the coats in which no one gets her coat back, and let $\tau$ be a permutation of the cases in which no one gets her case back. Then the ordered pair $(\sigma, \tau)$ is a way that no one gets her full possessions back.

There are $D_{10}$ choices for $\sigma$, and for each choice there are $D_{10}$ choices for $\tau$.

As to the second question, if everyone gets her coat back, and no one gets her case back, that would certainly not be part of the count for (a). But it would be part of the count for (b).

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thanks andre, one more quick question why is the answer to the second one $0$. Why cant no one women get back none of her possession. Wouldn't we treat both the possessions as one and the answer just be $d_{10}$ but in this instance it is $n!*S(2,10)$ –  Kj Tada Jul 16 '13 at 1:42
    
The answer is definitely not $0$. For example, if we give to person $i+1$ the possessions of person $i$ (and to $1$ the possessions of $10$, then no one gets back both of her possessions. And there are many other ways! –  André Nicolas Jul 16 '13 at 1:56

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