## Solution to puzzle of white tailed cats

The following solution is reproduced from Prof Ian Stewart’s book “Cabinet of Mathematical Curiosities”.

Suppose that there are c cats, of which w have white tails. There are $c(c-1)$ ordered pairs of distinct cats, and $w(w-1)$ ordered pairs of white tailed cats. (You can choose the first cat of the pair in c ways, but the second in only $c-1$ ways since you have used up one cat. Ditto for white tailed cats. By ordered, what is meant is choosing first cat A and then cat B is considered to be different from B first and then A. If you don’t like that, then both formulae have to be halved — with the same result.)
This means that the probability of both cats having white tails is $\frac{w(w-1)}{c(c-1)}$ and this must be 1/2. Therefore, $c(c-1)=w(w-1)$ with c and w being whole numbers. The smallest solution is $c=4$, and $w=3$. The next smallest turns out to  be $c=21$ and $w=15$. Since Ms Smith has fewer than 20 cats, she must have 4 cats, of which three have white tails.