Question

in how many ways can we divide 5 different coloured balls to three persons such that everyone must get atleast one ball..

Answer 1

Refer http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29
and get your answer

Answer 2

but that one will apply only to identical balls..

Answer 3

Similar problem asked by me http://openstudy.com/study#/updates/4f95edc7e4b000ae9ecbd6eb

Answer 4

First give each person a ball. This can be done in 5P3 ways.
Then distribute the remaining two balls to the 3 persons. This can be done is \(\large \binom{3}{2}\cdot 2\) ways.
Not entirely sure, though. If there are any errors in the reasoning, feel free to point them out.

Answer 5

Oops. Read stirling number of 2nd kind to get your solution

Answer 6

Sorry @quarkine for telling you to refer Star and bar Combinatorics

Answer 7

@shivam_bhalla think maybe i can couple the star one with 5! to get the ans...@blockcolder :lemme check ...

Answer 8

@quarkine: S(5,3)

Answer 9

@blockcolder :i think you havent considered the possibility that the remaining all two balls can be given to just one of the three..

Answer 10

You will also need 3!. now you can join them somehow to get the answer ;)

Answer 11

it looks like the balls can not be less than 1

in this case there are 5 ways in which one person can get balls.

only 1 ball . that means 5 ways for a ball, then (5+ 5p2+5p3+5p4+5factorial) This is for one person.

multiply it by 3

haha i am getting 975 ways.

Answer 12

Number of ways to assign 3 balls to person A, and 1 each to persons B and C
\[\left(\begin{matrix}5 \\ 3\end{matrix}\right)\left(\begin{matrix}2 \\ 1\end{matrix}\right)\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]
There are 3C1 ways to choose the person to get the 3 balls
So there are 10*2*1* 3= 60 ways for one person to have 3 balls and the other 2 have 1 ball.

There are
\[\left(\begin{matrix}5 \\ 2\end{matrix}\right)\left(\begin{matrix}3 \\ 2\end{matrix}\right)\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]
ways for person A to have 2 balls, person B to have 2 balls and C to have 1.
There are 3C1 ways to choose the person to get 1 ball. So
10*3*1 *3 = 90 ways for 2,2,1

Total number of ways= 150

Answer 13

@FoolForMath :no i dont hav the answer...

Answer 14

i think it may be 5P3, as @blockcolder said, times 3X3(which would take care of the possible ways to distribute the remaining 2) so it should be 540..

Answer 15

anyone got the answer to this??

Answer 16

Have a look at @phi 's answer