Unformatted text preview: N (2) = N (0) + 3 N (1) + 3 N (2) = 1 + 3(1) + 3(3) = 13 . We also ±nd for N (4) that N (4) = 3 ± j =0 ² 4 j ³ N ( j ) = ² 4 ³ N (0) + ² 4 1 ³ N (1) + ² 4 2 ³ N (2) + ² 4 3 ³ N (3) = N (0) + 4 N (1) + 3 · 4 2 N (2) + 4 N (3) = 1 + 4(1) + 6(3) + 4(13) = 75 . Problem 17 (why the binomial equals the multinomial) The expression ² n r ³ is the number of ways to choose r objects from n , leaving another group of nr objects. The expression ² n r, nr ³ is the number of divisions of n distinct...
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 Winter '08
 G.JOGESHBABU
 Algebra, Probability, person, Row, 0 3 j, 0 4 j

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