# Principle of I/E: In how many ways can eight cakes be distributed among four kids if every kid receives at least one cake?

Eight cakes are distributed randomly among four kids. Use I/E to determine in how many of the possible distributions every kid receives at least $$1$$ cake.

Hint: Define $$A_i$$ to be the set of distributions where the $$i$$th kid does not receive any cake. Compute $$|\cap_{i \in I} A_i|$$ if $$|I| = k \leq 4$$.

• The problem is asking you to subtract those distributions in which one or more children do not receive a cake from the total number of distributions. For instance, $A_1$ is the set of distributions in which the first child does not receive a cake, meaning that the eight cakes are distributed to the remaining three children. You should start the problem by determining in how many ways eight cakes could be distributed to four children without restriction. – N. F. Taussig Feb 18 at 1:23
• @N.F.Taussig do you know a formula to solve it? I don't get it – Unknown X Feb 18 at 1:33
• The Inclusion-Exclusion Principle for four sets is \begin{align*}|A_1 \cup A_2 \cup A_3 \cup A_4| & = |A_1| + |A_2| + |A_3| + |A_4| \\ & \quad - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_1 \cap A_4| - |A_2 \cap A_3| - |A_2 \cap A_4| - |A_3 \cap A_4|\\ & \quad + |A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_4|\\ & \quad - |A_1 \cap A_2 \cap A_3 \cap A_4|\end{align*} You need to compute $|A_1 \cup A_2 \cup A_3 \cup A_4|$ and subtract it from the total number of ways of distributing eight cakes to four children without restriction. – N. F. Taussig Feb 18 at 1:43
• without restriction I need to do a binomial coefficient right? @N.F.Taussig – Unknown X Feb 18 at 1:46
• If there are no restrictions, there are four possible recipients for each of the eight cakes, so how many distributions are possible? – N. F. Taussig Feb 18 at 1:47