It is equivalent to arrange the following 20 symbols, A1, A2, ..., A10www.ddhw.com B1, B2, ..., B10 under some rules. (1) they can occupy same position. (2) B1 must behind A1, B2 hehind A2, and so on. We can first ignore the 2nd reqirement, but use instead the condition that A1 and B1 cannot be in the same position, and so on. The result divided by 2^10 is what we need. we have such recurrence relationship, T(n+1, 2) = T(n, 2) * P(2,2)www.ddhw.com T(n+1, 3) = T(n, 2) * 2*3 +T(n, 3) *P(3,2) T(n+1, 4) = T(n, 2) *3*4 +T(n, 3) * 3*4 +T(n, 4) * P(4,2) ... T(n+1, m) = T(n, m-2) *(m-1)*m +T(n, m-1) *(m-1) *m +T(n, m)*P(m, 2) where T(n, m) denotes n pairs occupy m positions. We can use matrix multiplication to present the above relations. A computer program will do the calculations. I don't find a general expression for this problem. www.ddhw.com
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