Trillian's mice II

Assume the other mice are black, because black mice are da bomb (plz NSA). Denote the number of black mice as b. Using the given, we have $b \choose 2$ $2 \choose 2$ $=$ $b \choose 4$ Since Pascal's triangle is symmetric, b is 6. 6+2 = $\boxed{8}$

nC2=nC4 according to question where n is no. of non-white mice as in first case 2 places are already filled with white mice therefore probability nC2/(n+2)C4 and in other case we have to choose all four out of n non-white i.e. nC4/(n+2)C4. and thus nC2=nC4 or n=6 and answer is 6+2=8.

Let $x$ denote the total number of Trillian's mice. Then the probability of choosing two white mice is $\frac{{4 \choose 2}(2)(1)(x-2)(x-3)}{x(x-1)(x-2)(x-3)}$ , and the probability of choosing no white mice is $\frac{(x-2)(x-3)(x-4)(x-5)}{x(x-1)(x-2)(x-3)}$ . Setting the two expressions equal and simplifying, we get $x^2-9x+8=0$ , which has the solutions $x=1$ and $x=8$ . Since it is given that there are $2$ white mice, we know that the answer cannot be $1$ , so it is $\fbox{8}$ .