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

(A) 75 (B) 150 (C) 210 (D) 243

Lets assume the 3 people as A, B and C.

1st case  : A get 3 balls, both B and C get 1 ball. for this case, number of ways = 5C3*2C1*1C1= 20. which means that among 5 balls, choose 3 for A, then among the remaining 2 balls choose 1 for B, remaining 1 ball for C. And since it can be either A or B or C gets 3 balls. so total number of ways for this case= 20x 3=60

2nd case : A gets 2 balls, B gets 2 balls, C gets 1 ball. number of ways = 5C2*3C2*1C1=30. which means that among 5 balls, choose 2 for A ; among the remaining 3 balls choose 2 for B, and remaining 1 ball for C. also the same thing it can be A B get 2 balls, A C get 2 balls, B C get 2 balls. so total number of ways = 3x30=90

So summarize for both cases total number of ways= 90 + 60 = 150