Activity Building

Solution

We have \(15\) distinct participants and \(5\) distinct groups, with the condition that every group must have at least one person. Since the groups correspond to different activities, they are labeled. If there were no restriction, each of the \(15\) participants could go into any of the \(5\) groups, so the total number of assignments would be \[ 5^{15}. \] Now we remove the assignments where one or more groups are empty. Using inclusion–exclusion: \[ \text{Number of onto assignments} = 5^{15} -\binom{5}{1}4^{15} +\binom{5}{2}3^{15} -\binom{5}{3}2^{15} +\binom{5}{4}1^{15}. \] This is because: \[ \binom{5}{1}4^{15} \] counts assignments where at least one specified group is empty, \[ \binom{5}{2}3^{15} \] adds back assignments where at least two specified groups are empty, and so on. Now compute each term: \[ 5^{15}=30{,}517{,}578{,}125 \] \[ 4^{15}=1{,}073{,}741{,}824 \] \[ 3^{15}=14{,}348{,}907 \] \[ 2^{15}=32{,}768 \] So \[ 5^{15} -\binom{5}{1}4^{15} +\binom{5}{2}3^{15} -\binom{5}{3}2^{15} +\binom{5}{4}1 \] becomes \[ 30{,}517{,}578{,}125 -5(1{,}073{,}741{,}824) +10(14{,}348{,}907) -10(32{,}768) +5. \] So the total is \[ 30{,}517{,}578{,}125 -5{,}368{,}709{,}120 +143{,}489{,}070 -327{,}680 +5. \] This gives \[ 25{,}292{,}030{,}400. \] Therefore, the number of ways to divide the \(15\) participants into \(5\) nonempty labeled groups is \[ 25{,}292{,}030{,}400. \]