2nd Mexico 1988 Problem A1 In how many ways can we arrange 7 white balls and 5 black balls in a line so that there is at least one white ball between any two black balls?
56 Solution We have BWBWBWBWB and 3 spare W which can go anywhere. There are 6 different places to put each W (at either end or with one of the existing W). If we put each W in a different place then there are 6·5·4/3! = 20 possibilities. If we put two in one place and one in another, there are 6·5 = 30 possibilities. If we put them all in the same place, there are 6 possibilities. Total 56. As a check, put down the 7 W first. Then there are 8 places to put the five B, and each must go in a different place. Hence 8·7·6·5·4/5! = 56. 2nd Mexican 1988 © John Scholes jscholes@kalva.demon.co.uk 21 February 2004 Last corrected/updated 21 Feb 04