Six ice hockey sticks

Of the $6!$ possible assignments of sticks to players, we want to count those which give 1 stick to the correct player, and none of the other 5 sticks to the correct player.

There are $6$ ways to choose the one correct assignment. Using derangements , there are $D_5 = 44$ ways to give the other sticks to incorrect players.

The probability is thus $\frac{6 \cdot 44}{6!} = \frac{44}{120} = \frac{11}{30},$ so $a+b = \boxed{41}.$

This question is a good example for derangement. First, we can select one person out of 6 who gets his own stick in 6 ways.

Total sample space = 6!

5 persons stick can be disarranged in 44 ways is based on Derangement Formula link.

Hence, the required probability is,

44*6/6! = 11/30 = a/b.
a+b = 11+30 = 41.

Total no.of ways = $6!$ because 6 sticks can be arranged among 6 persons in this many ways.

Now, we choose and fix a person who gets his own stick (which can be done in $6$ ways).

Rest of the 5 persons won't get their stick. Hence, the no. of derangements possible $=5! \times (1 - \frac{1}{1!} + \frac{1}{2!} -\frac{1}{3!} +\frac{1}{4!} -\frac{1}{5!}) = 5! \times \frac{11}{30}$

So, no. of ways in which exactly one player picks his own stick = $6 \times 5! \times \frac{11}{30} = 6! \times \frac{11}{30}$

Thus, required probability = $\frac{6! \times \frac{11}{30}}{6!} = \frac{11}{30}$

And, of course, $11 + 30 = 41$