permutation

[nivas]

There are 5 balls of different colours and 5 boxes of colours the same as those of the balls. Number of ways in which the balls, one in each box can be placed such that a ball does not go to a box of its own colour is

[Georgi Stanoev]

i think it is 64

[Tristar]

There are 5 balls of different colours and 5 boxes of colours the same as those of the balls. Number of ways in which the balls, one in each box can be placed such that a ball does not go to a box of its own colour is?

 

Ans:

 

I solved this problem by the way of making generalizations. If there are just three elements (balls) 1, 2, 3 and three corresponding boxes B1, B2 and B3, then the total number of ways in which each ball can be placed into any one of the boxes B1, B2 or B3 will be given by: 3! = 6 arrangements. These arrangements are as follows: (123), (132), (213), (231), (312) and (321). Out of these 6 arrangements the first two are eliminated outright (as the first ball cannot go into the first position=B1). Out of the remaining 4 outcomes half satisfy the condition and the other half doesn’t. So, the number of ways for arranging 3 balls into 3 boxes such that a ball does not go into a box of its own color will be 2.

 

If we do it for 5 balls and five boxes, then the solution is 48 arrangements (5! = 120 total arrangements, (120/5) = 24 arrangements of 5 elements and we can subdivide 120 arrangements into five columns of 24 arrangements. Then, 120 – 24 = 96 (we eliminate all arrangements starting with 1 placed on the first position out of five). For the remaining 4 columns of 24 arrangements half satisfy our required condition and the other half does not. So, we have: 12 * 4 = 48 arrangements.