derangements basic practice question 2

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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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 $\tau$.
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