2006 Canadian MO Problems/Problem 1
Let be the number of ways distributing candies to children so that each child receives at most two candies. For example, , , and . Evaluate .
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|2006 Canadian MO (Problems)|
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The number of ways of distributing k candies to 2006 children is equal to the number of ways of distributing 0 to a particular child and k to the rest, plus the number of ways of distributing 1 to the particular child and k ¡ 1 to the rest, plus the number of ways of distributing 2 to the particular child and k ¡ 2 to the rest. Thus f(2006; k) = f(2005; k) + f(2005; k ¡ 1) + f(2005; k ¡ 2), so that the required sum is 1 + 1003 X k=1 f(2005; k) : In evaluating f(n; k), suppose that there are r children who receive 2 candies; these r children can be chosen in ¡n r ¢ ways. Then there are k ¡ 2r candies from which at most one is given to each of n ¡ r children. Hence f(n; k) = b X k=2c r=0 µ n r ¶µ n ¡ r k ¡ 2r ¶ = X1 r=0 µ n r ¶µ n ¡ r k ¡ 2r ¶
with ¡ x y ¢ = 0 when x < y and when y < 0. The answer is 1003 X k=0 X1 r=0 µ 2005 r ¶µ2005 ¡ r k ¡ 2r ¶ = X1 r=0 µ 2005 r ¶ 1003 X k=0 µ 2005 ¡ r k ¡ 2r ¶