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Divide 7 candies and 14 cakes among 4 children, requiring that each child gets at least one candy and that each child gets more cakes than candies. How many ways are there to do this? Candies are indistingushable, cakes are indistinguishable, children are distinguishable

Is this right? Since each child get at least one candy than 7- 4 =3. So

$$c_1 + c_2 + c_3 + c_4 = 3$$

And more cake than candy so 14 -8= 6. So $$w_1 + w_2 + w_3 + w_4 = 6$$

Then $$\binom{3 + 4 - 1}{4 - 1} \cdot \binom{6 + 4 - 1}{4 - 1}$$

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  • Start by giving each child one candy and two cakes. Each child has a candy and more cake than candy.
  • Pair the three remaining candies with a cake apiece. Using the stars and bars formula, you can distribute these pairs to the four children in $6\choose 3$ ways. Each child still has more cake than candy.
  • There are 3 remaining cakes to be handed out, which again can be given to the children in $6\choose 3$ ways.

That's a total of $20\cdot20=400$ possible combinations.

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