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How many way can we distribute 9 balls numbered from 1 to 9 in 9 boxes numbered from 1 to 9 such that each box contain exactly one ball and either of the balls with number 1 2 3 are not in boxes 1 2 3. Please help me solve the problem by not using n(AUBUC). Probably by using dearrangments.
Hi Jayavignesh,
The solution to this question is as follows:
Total number of ways of distributing 9 balls = 9!.
Number of ways in which ball 1 is placed in box 1 = 8!
Number of ways in which ball 2 is placed in box 2 = 8!
Number of ways in which ball 3 is placed in box 3 = 8!
Number of ways in which balls 1 and 2 are placed in box 1 and 2 = 7!
Number of ways in which balls 2 and 3 are placed in box 2 and 3 = 7!
Number of ways in which balls 1 and 3 are placed in box 1 and 3 = 7!
Number of ways in which balls 1, 2 and 3 are placed in box 1,2,3 = 6!
Total number of ways = 9!-(8!+8!+8!-7!-7!-7!+61) = 9!-3*8!+3*7!-6!
The dearrangements cannot be applied in this question because we have to apply dearrangements formula many times to arrive at the solution, Balls 1,2,3 cannot be placed in boxes 1,2,3 but balls 4-9 can be placed in boxes 4-9. We have to isolate each case for boxes 4-9 and then apply the formula which would be very painful
There are 9 balls and 9 boxes.
Number of ways of arranging 9 balls in 9 boxes = 9!
We will subtract the required cases:
AUBUC = A+B+C-AintB-BintC-CintA+AintBintC
Let A be when only ball 1 is inserted in box 1 = 8!
Let B be when only ball 2 is inserted in box 2 = 8!
Let C be when only ball 3 is inserted in box 3 = 8!
AintB = When both balls 1 and 2 are inserted in boxes 1 and 2 = 7!
BintC = When both balls 2 and 3 are inserted in boxes 2 and 3 = 7!
AintC = When both balls 1 and 3 are inserted in boxes 1 and 3 = 7!
AintBintC = When balls 1,2,3 are inserted in boxes 1,2,3 respectively = 6!
AUBUC = 3*8!-3*7!+6!
Required number of ways = 9!-(3*8!-3*7!+6!)
