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Priyesh (1601)

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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

"Imagination is more important than knowledge."

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