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The total no. of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets atleast one ball is ?

$\begin{array}{1 1} 75 \\ 150 \\ 210 \\ 243 \end{array}$

1 Answer

Total no. of balls $=5$
No. of persons $=3$
Each person should get atleast one ball.
$\therefore$ The selection of balls can be done by
$(1,1,3)$ or $ (2,2,1)$
$i.e., \:(^5C_1\times.^4C_1\times^3C_3) \:\:or\:\:(^5C_2\times^3C_2\times^1C_1)$
$=(5\times 4\times 1)\:\:or\:\:(10\times 3\times 1)$
$=(20)\:\:or\:\:(30)\:\:ways.$
After selection it should be distributed between three different persons.
$(1,1,3)$ selection is distributed in $\large\frac{3!}{2!}$$=3$ $ways\:\:and$
$(2,2,1)$ selection is distributed in $\large\frac{3!}{2!}$$=3$ $ways.$
$\therefore$ The required no. of ways $=(20\times 3)+(30\times 3)$
$=60+90=150$
answered Aug 12, 2013 by rvidyagovindarajan_1
 

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