(144)

Sol. The given equation is x + y + z = 20 .........

We have to find the number of different values of x, y, z

Such that x ≠ y ≠ z and x, y, z ≥ 1

Let us assume that x < y < z

And x = x 1 , y – x = x 2 and z – y = x 3

Then x = x 1 ; y = x 1 + x 2 and z = x 1 + x 2 + x 3

Also x 1 , x 2 , x 3 ≥ 1

Substitution these values in we get

3x 1 + 2x 2 + x 3 = 20 ..........

Where x 1 , x 2 , x 3 ≥ 1

Now No. of solution of equation is

= Co-efficient of x 20 in (x 3 + x 6 + x 9 + ....) × (x 2 + x 4 + x 6 + ....) × (x + x 2 + x 3 + ...)

= Co-efficient of x 14 in (1 + x 3 + x 6 + x 9 + ....) (1 + x 2 + x 4 + -....) × (1 + x + x 2 +....)

= (1 + x 2 + x 3 + x 4 + x 5 + 2x 6 + x 7 + 2x 8 + 2x 9 +2x 10 + 2x 11 + 3x 12 + 2x 13 + 3x 14 + 3x 15 + ....)

(1 + x + x 2 + x 3 + .......)

Co-efficient of x 14 is 1 + 1 + 1 + 1 +1 + 2 +1 + 2 + 2 + 2 + 2 + 3 + 2 + 3 = 24

But x , y and z are arranged in 3! ways

So Required no of solution = 24 × 6 = 144 Ans.

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