1. Rewrite the generic term as C(m,m-k)*C(n,r+k). In this form the target, C(m+r,m+n) accounts for all paths to location m+r at depth m+n. The lead term, C(m,m)*C(n,r) says do m of the m+r right terms at level m, then drop down the remaining n levels via all paths with r right turns; the next term C(m,m-1)*C(n,r+1), says do one fewer right turn up to depth m, then add one extra in the second half drop to level m+n. Every path through level m is accounted for in this way, each with the appropriate ending paths to get to C(m+r,m+n).

2. Diferentiate (1-x)^{n} and its binomial expansion (then set x = 1).

3. Rewrite general term as C(n,n-k)*C(m,k). Given a set of m+n things, count size n subsets, beginning with all the n things, then one n thing missing, + one chosen m thing, then 2 n things missing, + 2 chosen m things, and so forth. So: C(m+n,n).

4. C(23,15)- C(9,1)*C(18,10)+C(9,2)*C(13,5)-C(9,3)

5. Coefficient of x^{r} in (1+x+x^{2}+....+x^{k} +...)^{5}

6. Fourth expression captures a; third expression captures b-a; and so forth.

7. This is x^{35}*(1+x+x^{2}...)^{7}; coefficient of x^{(r-35)} is C(r-41,r-35).

8. (1+x+x^{2}..+x^{9})^{5}, coefficient of x^{r}

9. Treat as 10 balls in 5 boxes, each box gets 0 - 5 balls. Seek coefficient of x^{10} in (1+x+x^{2}..+x^{5})^{5 }

10. (1+x+x^{2}+x^{3})^{3}(1+x+x^{2}+....+x^{k} +...)^{2} seek coefficient of x^{10}.

11. Think of this as 1/(1-u), where u is (x+x^{2}+....+x^{6})

-- RobbieMoll - 2012-05-01

Topic revision: r3 - 2012-05-03 - RobbieMoll

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