Mathematics - Permutation & Combination Question with Solution | TestHub

$\because$ Each person gets at least one ball. $\therefore 3$ Persons can have 5 balls as follow. \begin{array}{|c|c|c|} \hline Person & No. of balls & No. of balls \\ \hline I & 1 & 1 \\ \hline II & 1 & 2 \\ \hline III & 3 & 2 \\ \hline \end{array} The number of ways to distribute balls 1,1,3 in first to three persons $={}^5{\mathrm{C}}_1\times {}^4{\mathrm{C}}_1\times {}^3{\mathrm{C}}_3$ Also 3 , persons having 1,1 and 3 balls can be arranged in $\frac{3!}{2!}$ ways. $\therefore$ Total no. of ways to distribute $1,1,3$ balls to the three persons $={}^5{\mathrm{C}}_1\times {}^4{\mathrm{C}}_1\times {}^3{\mathrm{C}}_3\times \frac{3!}{2!}=60$ Similarly, total no. of ways to distribute $1,2,2$ balls to three persons $={}^5{\mathrm{C}}_1\times {}^4{\mathrm{C}}_2\times {}^2{\mathrm{C}}_2\times \frac{3!}{2!}=90$ $\therefore$ The required number of ways $=60+90=150$