By Ross S., Weatherwax J.L.

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**Example text**

Then since all paintings are sold we have the following constraints on Di , Gi , and Pi , 5 5 Di = 4 , 5 Gi = 5 , i=1 i=1 Pi = 6 . i=1 Along with the requirements that Di ≥ 0, Gi ≥ 0, and Pi ≥ 0. Remembering that the number of solutions to an equation like x1 + x2 + · + xr = n , n+r−1 . Thus the number of solutions to the first equation r−1 8 4+5−1 = 70, the number of solutions to the second = above is given by 4 5−1 9 5+5−1 = 126, and finally the number of solutions to = equation is given by 4 5−1 6+5−1 10 the third equation is given by = = 210.

2 So in total we have by summing up all these mutually exclusive events we find that the total number of five digit numbers allowing repeated digits is given by 9·8·7·6·5+9 5 2 ·8·7·6+ 1 ·9· 2 5 2 8 3 2 · 7 = 52920 . Problem 11 (counting first round winners) Lets consider a simple case first and then generalize this result. Consider some symbolic players denoted by A, B, C, D, E, F . Then we can construct a pairing of players by first selecting three players and then ordering the remaining three players with respect to the first chosen three.

Once both of these classes are selected we pick the individual two and one students from their n n ways respectively. Thus in total we have and respective classes in 1 2 3·2· n 2 n 1 = 6n n(n − 1) = 3n2 (n − 1) , 2 ways. Part (d): Three students (all from a different class) can be picked in Part (e): As an identity we have then that 3n 3 =3 n 3 + 3n2 (n − 1) + n3 . n 1 3 = n3 ways. We can check that this expression is correct by expanding each side. Expanding the left hand side we find that 3n 3 = 3n(3n − 1)(3n − 2) 9n3 9n2 3n!