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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, 10 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. Subharthi Chowdhuri · 3 years, 10 months ago

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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. Tim Vermeulen · 3 years, 10 months ago

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3^5 =243. (D)is the answer. Bhargav Das · 3 years, 10 months ago

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@Bhargav Das how to do it???? Rajath Krishna R · 3 years, 10 months ago

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@Rajath Krishna R 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

\( 3*3*3*3*3 = 3^{5} = 243 \) Luca Bernardelli · 3 years, 10 months ago

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@Luca Bernardelli Note that every player gets at least one ball. Tim Vermeulen · 3 years, 10 months ago

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