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The number of ways in which 26 identical chocolates be distributed between Amy, Bob, Cathy and Daniel so that each receives at least one chocolate and Amy receives more chocolates than Bob is
Correct Answer: 1078
The total number of positive solutions of $$x_1+x_2+x_3+x_4=26$$ is $$\binom{25}{3}=2300$$. The cases with $$x_1=x_2$$ satisfy $$2x_1+x_3+x_4=26$$ and give $$22+20+\cdots+2=144$$ solutions. By symmetry, half of the remaining solutions have $$x_1>x_2$$, so the answer is $$\frac{2300-144}{2}=1078$$.
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