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

This is a case of distinct to distinct.

Case 1. groups of 1, 1, 3 gives ${5 \choose 1}$ ${4 \choose 1}$ ${3\choose 3}$= 20

Case 2 groups of 1, 2, 2 gives ${5 \choose 1}$ ${4 \choose 2}$ ${2 \choose 2}$ = 30

Now here my problem arises.

Take case 1

Assume balls to be A B C D E. You make groups A, B, CDE.

Let persons be 1 2 3.

The following distributions can be done (one column is one distribution)

PERSON 1   A        A      B       B      CDE       CDE

PERSON 2   B       CDE     A      CDE      A         B


PERSON 3  CDE       B      CDE      A      B         A

That is in case 1, each group made can be distributed in 6 ways (3!). Therefore 20 groups can be distributed in 20 x 6 = 120 ways.

Similarly for case 2 we get 30 x 6 = 180 ways.

Therefore total is equal to 600 ways.

But the answer given is 150. Each distribution is taken as 3!/2!. Why is that ? The question specifies different colour balls and to person and persons are always different.

Shaurya Goyal
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  • Alternative approach: paint on every ball the $0$-based number of person to whom the ball goes, place them in a row, there will be $3^5$ ways if removing constraints. Now we say wlog $2$nd person receives no balls, this may happen in $2^5$ ways, and we're overcounting the cases when a person get all the balls $2$ times, so the answer is $3^5-3\cdot 2^5+3=150$ – Alexey Burdin Jul 24 '20 at 15:20
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    Case 1 and 2 are independent of each other so you've to add them instead of multiplying and you get $300$. Also, two of the groups of each case are of equal size so divide by $2!$. The final answer is $150$. – SarGe Jul 24 '20 at 15:22
  • First think the simple case: what about only 2 persons. Since each ball can go to any person, there are $2^5-2$ cases, where subtracting 2 is to exclude cases all 5 balls go to a single person. Therefore, the answer to 3 persons is $3^5 -{3 \choose 2}(2^5-2)-3$, where 3 is the number of cases that all 5 balls go to a single person. – toronto hrb Jul 24 '20 at 15:29

2 Answers2

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You can distribute 5 balls with at least one ball to each person in 6 ways with the distribution as follows: (1,1,3), (1,2,2),(2,1,2),(2,2,1),(1,3,1),(3,1,1) In case 1, person 1 gets 1 ball, person 2 gets 1 ball and person 3 gets the remaining. In case 2, person 1 gets 1 ball, person 2 gets 2 and person 3 gets remaining 2. I hope you can understand the rest of the cases also. Now, case 1 can happen in $ {5 \choose 1} ×{ 4\choose 1} × {3\choose 3 }= 20 $ ways. Case 2 can happen in${ 5 \choose 1}× {4 \choose 2}× {2\choose 2} = 30 $ ways.
Similarly, by applying suitable selection formula for other cases with the help of the distribution given, case 3 gives $30$, case 5 gives $30$, case 5 gives $20$ and case 6 gives$ 20$ ways. Summing up, we get 150.

Laxmi
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  • Use standard notation instead of $nCr$ which is $\displaystyle{n\choose r}$ by typing n\choose r. – SarGe Jul 24 '20 at 15:43
  • Thanks, I was trying to find the code for nCr only. I hope it's correct now. Could you also suggest me how to begin in a new line? – Laxmi Jul 24 '20 at 16:11
  • In mathmode use \\ while in textmode you can use <br>. Also, to start new paragraph press Enter key twice on keyboard. – SarGe Jul 24 '20 at 17:53
  • Noted. Thanks a lot. – Laxmi Jul 25 '20 at 14:10
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Hint: if you simply multiply with $3!$, then $6$ permutation of $A,B,CDE$ are different from $6$ permutation of $B,A,CDE$

similar logic to $1,2,2$

A neat solution is using PIE: $3^{5}-3\times 2^{5}+3\times 1^{5}=150$

acat3
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