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There are 4 balls of different colour and 4 boxes of colours, same as that of balls. The no. of ways in which the balls, one each in a box,could be placed such that a ball does not go to a box of its own colour is ?

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Let $B_1,B_2,B_3,B_4$ be balls and
$A_1,A_2,A_3,A_4$ be the respective boxes of respective colour.
Let ball $B_1 $ does not go in box $A_1$ which is of same colour of that of the ball.
$i.e.,B_1$ can be placed in any of the three boxes $A_1,A_2,A_3$.
If $B_1$ is placed in box $A_2$, then the other 3 balls can be placed in the following 3 ways.
$A_1$ $B_2$ $B_3$ $B_4$
$A_3$ $B_4$ $B_4$ $B_2$
$A_4$ $B_3$ $B_2$ $B_3$

 

Similarly if$B_1$ is placed in box $A_3$ or in $A_4$ we get 3 ways in each.
$\therefore$ The required no. of ways = $3 \times 3=9$

 

answered Aug 31, 2013 by rvidyagovindarajan_1
edited Aug 31, 2013 by rvidyagovindarajan_1
 

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