Let A= 1,2,3,4,5,6,7,8 , B= 9,10,11,12 , 13,14,15,16 and C= 17,18,19,20,21,22,23 , 24 . Find the number of triples (x, y, z) such that x A, y B, z C and x+y+z=36 .

Step-by-step solution

x, y, z N and array ll x A, i.e., & 1 x 8 and y B, i.e., & 9 y 16 and z C, i.e., & 17 z 24 array Number of solution for x+y+z=36 is equal to coefficient of x³⁶ in (x+x^2+ +x^8 ) . (x^9+x¹⁰+ + . .x¹⁶ ) (x¹⁷+x¹⁸+ +x²⁴ )= coefficient of x^9 in (1+x+ +x^7 )^3= coefficient of x^9 in (1-x^8 )^3 (1-x)^3 = coefficient of x^9 in (1-3 x^8+3 x¹⁶-x²⁴ ) aligned & (1+3 x+ 3.4 2! x^2+ ) = & ¹¹ C₉-3.3 = & 55-9 = & 46 aligned