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Hence after filling 1st box in one way we can fill the second and third boxes in 10 different box.
1st box itself can be filled in 10 different ways.
Hence the total number of ways the boxes can be filled is 100 ways. (one hundred ways.)
I think this is the right answer
Therefore total no of ways of distribution of the 12 identical balls in the 3 identical boxes=(10+9)*1=19
since the balls & boxes are identical so we are not bothered about which ball goes into which box.We are only going to consider the no. of balls going into each box. & since boxes are identical hence arrangements such as (6,5,1) & (5.6.1) will be same
now suppose no. of balls going into first,second & third boxes are x1, x2, x3 respectively
then x1 + x2 +x3 = 12
& furthermore (x1,x2,x3) this pair of three should not rearrange
so first total no. of integral solns. = 14C2 = 13*7 = 91
no when x1=x2=x3 we have have one soln i.e (4,4,4)
when x1=x2!=x3 we have
(0,0,12)(1,1,10),(2,2,8),(3,3,6),(5,5,2),(6,6,0) i.e 6 solns
when x1!= x2!=x3
[91 - 1 - 6 * (3!/2!)]/3! = (91-19)/6 = 72/6 = 12
so to verify 12* 3! + 6 * 3!/2! + 1 = 91
hence required answer is 12 + 6 + 1 = 19
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let three boxes get any no. of balls let say x,y,z. now the equation you can form is x + y + z = 12. the no. of non-negative roots of this eq. is 14C2.