The x , y , z x, y, z series

Probability Level 3

How many ordered triples of positive integers ( x , y , z ) (x,y,z) are there such that x + y + z = 20 x+y+z=20 and two of x , y , z x,y,z are odd?

138 141 135 145

Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

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1 solution

Grant Bulaong
Nov 16, 2016

Two of the numbers are odd so the last must be even. There are 3 3 ways to set which number is even.

Assume z z is the even number and let x = 2 a − 1 x=2a-1 , y = 2 b − 1 y=2b-1 and z = 2 c z=2c where a , b , c a,b,c are positive integers.

We solve for the number of solutions to the equation a + b + c = 11 a+b+c=11 . By Stars and Bars, we have 10 C 8 = 45 10C8=45 .

Permuting the number parities, the total number of solutions is ( 45 ) ( 3 ) = 135 (45)(3)=135 .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Fancy meeting you here :D

Another sol'n is by taking the # of positive triplets for which x + y + z = 20 x+y+z = 20

Then subtracting the # of positive triplets for which x + y + z = 10 x+y+z = 10 :D

Manuel Kahayon - 4 years, 5 months ago

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