[Copernicus]

Another way to reason it from where you got is:

Invert both sides of your equation and call B+W=T(otal) (just to make thinking about it easier) .

Then T*(T-1)*(T-2)*(T-3)*(T-4)/(B)*(B-1)*(B-2)*(B-3)(B-4)=2

For the ratio to be 2, if you factor all of the numbers in the N(umerator) and (Denominator) into their prime factors, everything has to cancel in the N and the D except 2*P in the N and P in the D.

Since all but one of the prime factors exist in both the N and D, the product of the non-unique prime factors in each the N and D must be the same. The constraints that the original numbers are consecutive in both N and D, and their products are the same means they must be the same numbers. (This is provable, but beyond this little post).

The problem therefore reduces to, "what is the smallest set of 5 consecutive numbers in the numerator that has 4 identical consecutive numbers in the denominator, except that the remaining number in the N is 2*the remaining number in the D?" Since the N is larger, you also know that its non-cancelling number is larger than the non-cancelling number in the D. Further, since the cancelling numbers were shown to be consecutive, the non-cancelling number in the N must be T, and the non-cancelling number in the D must be T-5 (because T-1, T-2, T-3, and T-4 are the only possible cancelling numbers in both the N and D). The problem therefore becomes
T/(T-5)=2, or T=10. Since there is only root to the equation, not only is T=10 the minimum but also the only solution.

Iteration is easier, and from the iteration you can probably prove by induction that there is no larger set of numbers that can meet the criteria. The above, however, is more rigorous and independent of having a "starting point" for an induction proof.