Question

The total number of ways in which 5 different balls can be distributed among 3 persons so that each person gets at least one ball is
A) 75
B) 150
C) 210
D) 243

Answer 1

There are five different balls and distributed among 3 persons such that each person gets at least one, then there are only two arrangements to distribute balls among people. The first one is 1,1,3 and second is 1,2,2. To find total ways just find the number of ways to arrange these two arrangements. There is also a short method, find all ways to distribute 5 balls and subtract the ways in which at least one person gets no balls.

Complete step-by-step answer:
There are two possible arrangements of balls 1,1,3 and 1,2,2 for distributed balls such that each person gets at least one ball.
For arrangement 1,1,3 number of ways to distribute balls are
$^5{C_1}{ \times ^4}{C_1}{ \times ^3}{C_3} = 5 \times 4 \times 1 = 20$ ways.
Also, we can distribute 1,1,3 as 1,3,1 and 3,1,1.
Then the total number of ways to distribute balls as 1,1,3 is 20x3=60.
For arrangement 1,2,2 number of ways to distribute balls are
$^5{C_1}{ \times ^4}{C_2}{ \times ^2}{C_2} = 5 \times 3 \times 2 = 30$ ways.
Also, we can distribute 1,2,2 as 2,1,2 and 2,2,1.
Then the total number of ways to distribute balls as 1,2,2 is 30x3=90.
So, the total number of ways to distribute balls such that each person gets at least one ball is 60+90+150.

Hence, the correct answer is option B.

Note: Total number of ways to distribute five different balls among 3 persons are ${3^5}$.
Number of ways to distribute five balls to 3 people at least one get no ball are $^3{C_1} \times ({2^5} - 1)$ ways.
Then total number of ways to distribute balls such that each person get at least one ball is
${3^5}{ - ^3}{C_1} \times ({2^5} - 1) = 243 - 3*31$
$= 243 - 93 = 150$