Unformatted text preview: the women that get oﬀ on ﬂoor i. Thus we must have
6 i=1 6 mi = 5 mi ≥ 0 wi = 3 wi ≥ 0 . i=1 The number of solutions to the ﬁrst equation is given by 5+6−1 6−1 3+6−1 6−1 = 10 5 = 252 , while the number of solutions to the second equation is given by = 8 5 = 56 . So in total then (since each number is exclusive) we have 252 · 56 = 14114 possible elevator situations. Problem 33 (possible investment strategies) Let xi be the number of investments made in opportunity i. Then we must have
4 xi = 20
i=1 with constraints that x1 ≥ 2, x2 ≥ 2, x3 ≥ 3, x4 ≥ 4. Writing this equation as x1 + x2 + x3 + x4 = 20 we can subtract the lower bound of each variable to get (x1 − 2) + (x2 − 2) + (x3 − 3) + (x4 − 4) = 20 − 2 − 2 − 3 − 4 = 9 . Then deﬁning v1 = x1 − 2, v2 = x2 − 2, v3 = x3 − 3, and v4 = x4 − 4, then our equation becomes v1 + v2 + v3 + v4 = 9, with the constraint that vi ≥ 0. The number of solutions to equations such as these is given by 9+4−1 4−1 = 12 3 = 220 . 4 = 4 possible 3 ways. The four choices are denoted in table 1, where a one denotes that we invest in that option. Then investment choice number one requires the equation v2 +v3 +v4 = 20−2−3−4 = Part (b): First we pick the three investments from the four possible in ...
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 Winter '08
 G.JOGESHBABU
 Probability, 5 mi, 1 6 mi, possible elevator situations

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