MA340 Study Guide - Negative Binomial Distribution, Poisson Point Process, Triangular Distribution

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3. 19(b) before we actually start the problem let"s get a feel for it in the case n = 5. To avoid confusion assume the boxes are labelled 1, 2, 3, 4, 5 and the balls labelled a, b, c, d, e. an outcome can be represented as a. 5 tuple such as (1, 1, 3, 2, 1), and we interpret this as a going to box. 1, b going to box 1, c going to box 3, d going to box 2 and e going to box 1. There are clearly 55 total outcomes, and it is worth noting that the outcome (1, 1, 3, 2, 1) will leave exactly two boxes (boxes 4 and 5) empty. Well its going to have to look something like (1, 3, 1, 2, 5). We see that two of the balls have to go in the same box and the other three have to go in di erent boxes.

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