No. of Solutions Pls!!!

Probability Level pending

Find the no. of different solutions of x+y+z=13 where x,y and z are natural no.s


The answer is 66.

Harshit Joshi by Harshit Joshi , 23, India

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1 solution

Keegan Lee
Oct 30, 2014

the solutions for y+z=n are (1,n-1), (2,n-2)..., (n-1,1). so there are n-1 solutions. every solution must have y+z=13-x so for every natural number x, there are 12-x solutions. meaning there are only solutions where 0<x<12. the number of solutions to the entire thing is the sum of solutions for each x from 1 to 11, or...

∑ i = 1 11 ( 12 − i ) = ∑ i = 1 11 12 − ∑ i = 1 11 i = 11 ∗ 12 − 11 ( 11 + 1 ) 2 = 66 \displaystyle \sum_{i=1}^{11} (12-i)=\displaystyle \sum_{i=1}^{11}12 -\displaystyle \sum_{i=1}^{11}i=11*12-\frac{11(11+1)}2=66

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