Answer :
There are 10 different ways to distribute eleven cakes among three kids such that each kid gets not less than two cakes and not more than four cakes.
To solve this problem, we can use a combinatorial approach. We'll distribute the cakes among the kids while satisfying the given conditions: each kid gets not less than two cakes and not more than four cakes.
Let's consider the possible cases:
Each kid gets 2 cakes: In this case, all three kids get 2 cakes each. This is one possible distribution.
One kid gets 2 cakes, and the other two get 4 cakes each: There are 3 ways to choose which kid gets 2 cakes. Once that choice is made, the other two kids will get 4 cakes each. So, there are 3 possible distributions in this case.
Two kids get 2 cakes, and one kid gets 5 cakes: There are 3 ways to choose which kid gets 5 cakes. Once that choice is made, the other two kids will get 2 cakes each. So, there are 3 possible distributions in this case.
One kid gets 3 cakes, and the other two get 4 cakes each: There are 3 ways to choose which kid gets 3 cakes. Once that choice is made, the other two kids will get 4 cakes each. So, there are 3 possible distributions in this case.
So, the total number of different ways to distribute the cakes is 1 + 3 + 3 + 3 = 10 ways.
Therefore, there are 10 different ways to distribute eleven cakes among three kids such that each kid gets not less than two cakes and not more than four cakes.
To learn more about distributions, here:
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