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# Combinatorics problem help.........

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

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

Note by Rajath Krishna R
3 years, 6 months ago

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answer is 150. in how many ways can you distribute 5 balls among 3 people such that atleast one of them gets 1? it is nothing but 3,1,1 and 2,2,1. so the desired answer will be (5!/3!)(3!/2!)+ (5!/ (2!)^2)(3!/2!) you need to multiply by 3!/2! because there are these many ways to arrange 3,1,1 or 2,2,1 . By evaluating the expression answer will come as 150. · 3 years, 5 months ago

First consider how you can distribute $$5$$ balls over $$3$$ players while every player gets at least $$1$$ ball, where all balls are equal and all players are equal. You could give one player $$3$$ balls and the other players both $$1$$ ball, or you could give one player $$1$$ ball and the other players both $$2$$ balls. If you work out these cases separately, you'll find the answer. · 3 years, 5 months ago

3^5 =243. (D)is the answer. · 3 years, 6 months ago

how to do it???? · 3 years, 6 months ago

Ball 1 could go to person A, or B, or C -> 3 choices

Ball 2 could go to person A, or B, or C -> 3 choices

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Ball 5 could go to person A, or B, or C -> 3 choices

$$3*3*3*3*3 = 3^{5} = 243$$ · 3 years, 5 months ago