The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

(b) : We can have the following distributions \l c c c c l B_1 B_2 B_3 [1em] Case I 1 1 3 -> (5! / 1! 1! 3!) * (1 / 2!) = 10 [1em] Case II 2 2 1 -> (5! / 2! 2! 1!) * (1 / 2!) = 15 [1em] Total number of ways =25 x 13=25 x 6=150

The total number of ways in which 5 balls of different ... | Permutations and Combinations (Mathematics)

(b) : We can have the following distributions

\begin{array}{l c c c c l} & B_1 & B_2 & B_3 & & \\[1em] \text{Case I} & 1 & 1 & 3 & \rightarrow & \frac{5!}{1! \, 1! \, 3!} \cdot \frac{1}{2!} = 10 \\[1em] \text{Case II} & 2 & 2 & 1 & \rightarrow & \frac{5!}{2! \, 2! \, 1!} \cdot \frac{1}{2!} = 15 \end{array} \\[1em]

Total number of ways

=25 \times 13=25 \times 6=150

Explanation

For (2 leq r leq n,binom{n}{r} + 2binom{n}{r - 1} + binom{n}{r - 2} =)

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