counting on $4$ pairs of gloves There are $4$ different pairs of gloves. $4$ right handed gloves are given randomly to four persons
then $4$ left handed gloves are given.how many ways are possible such that nobody gets the 
right (correct)pair of gloves.
$\underline{\bf{My\;\;Try}}::$ Let the $4$ pairs of gloves as $(a,A)\;\;,(b,B)\;\;,(c,C)\;\;,(d,D)$.
Then R.H.S gloves $(A,B,C,D)$ can be distributed as $ = 4!$ ways.
Now I Did not understand How can I distributed L.H.S gloves so that nobody get correct pair of gloves
Help Required.
Thanks
 A: You can (and should) look up Derangements. But the numbers here are so small that we can list and count. Let the right gloves be called $A$, $B$, $C$, $D$, and the matching left gloves be $a,b,c,d$. The right gloves can be distributed between the $4$ people in $4!$ ways. 
Once you have done this, line up the right glove holders in the order $A,B,C,D$ and distribute the left gloves at random. 
We count the ways in which we can distribute the left gloves so there is no matching pair. If there is no matching pair, left glove $a$ must go  to $B$, $C$, or $D$. We count the number of ways there is no match and $a$ goes to $B$, and multiply the result by $3$.
So $a$ goes to $B$. Maybe $b$ goes to $A$. Then $c$ must go to $D$, and $d$ to $C$. That gives $1$ way.
Maybe $b$ goes to one of $C$ or $D$. We will count the ways in which $b$ goes to $C$, and multiply by $2$. If $a$ goes to $B$, and $b$ to $C$, then $d$ must go to $A$, and $c$ to $D$. That gives $1$ way. Multiply by $2$, since $b$ could have gone to $D$.
We conclude that there is a total of $3$ ways in which $a$ can go to $B$. Multiply by $2$. We get $9$.
Thus for every one of the $4!$ ways to distribute the right gloves among the $4$ people, there are $9$ ways to distribute the left gloves so that no one is happy, for a total of $4!\cdot 9$.
