# Greg eats 5 different chocolates. Whatever he doesnt eat, he gives it to Sherlock. Whatever Sherlock doesnt eat, he gives it to John. In how many ways can all 3 of them share the chocolates, if...

Greg eats 5 different chocolates. Whatever he doesnt eat, he gives it to Sherlock. Whatever Sherlock doesnt eat, he gives it to John. In how many ways can all 3 of them share the chocolates, if each of them eats atleast one.

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### 1 Answer

We begin with 5 chocolates; Greg always goes first followed by Sherlock and then by John. Each person gets at least 1 chocolate. We also assume that all of the chocolates are consumed.

The possibilities are 3-1-1, 2-2-1, 2-1-2, 1-3-1, 1-2-2, 1-1-3 where 3-1-1 means Greg eats three, Sherlock eats 1 and John eats 1.

(1) 3-1-1 Greg can eat 3 candies out of 5 in 10 ways. ( We assume that the order the candies are consumed does not matter so use combinations; `_5C_3=10` ) Then Sherlock has 2 choices and John only 1.

There are 10*2*1=20 ways to distribute the candies 3-1-1.

(2) 2-2-1 Greg has 10 ways to choose 2 candies. ( `_5C_2=10` ). Now Sherlock has 3 ways to choose 2 candies from the remaining 3 and John has 1 way to choose.

There are 10*3*1=30 ways to distribute the candies 2-2-1

(3) 2-1-2 Again Greg has 10 ways. Sherlock has 3 ways to choose 1 candy from the remaining 3, and John has 1 way to choose the last candies.

There are 10*3*1 ways to distribute 2-1-2

(4) 1-3-1 Greg has 5 ways to choose 1 candy. Sherlock has 4 ways to choose 3 from the remaining 4, and John has one choice.

There are 5*4*1=20 ways to disrtibute 1-3-1

(5) 1-2-2 Greg has 5 choices. Sherlock has 6 ways to choose 2 from the remaining 4, while John has only the one choice.

There are 5*6*1=30 ways to distribute 1-2-2

(6) 1-1-3 Greg has 5 choices. Sherlock has 4 ways to choose 1 from the remaining 4, while John has 1 choice.

There are 5*4*1=20 ways to distribute 1-1-3

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There are 150 ways to distribute the 5 candies under the given conditions.

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** If all of the candies need not be distributed, the answer changes. There are new possibilities to be considered; in addition to those stated you would have 1-1-1,1-1-2,1-2-1,2-1-1.

1-1-1 5*4*3=60 ways

1-1-2 5*4*3=60 ways

1-2-1 5*6*2=60 ways

2-1-1 10*3*2=60 ways

The new total is 150+240=390 ways.