number of ways of selecting two numbers from 1 to 12 such that their sum is divisible by 3.

Dear Naveen

for a no to be divisible by 3 , possibilities are 3,6,9,15,18,21

if 2 no. are x and y , den

eqns are

x + y = 3 ,x + y = 6, x + y = 9, x + y = 12 , x + y = 15 , x + y = 18 , x + y = 21

for 1st eqn x,y wud be 1 & 2

for 2nd set is (2,4),(1,5)

for 3rd (1,8),(2,7),(3,6),(4,5)

for 4th (1,11),(2,10),(3,9),(4,8),(5,7),(6,6)

for 5th (12,3),(11,4),(9,6),(10,5),(8,7)

for 6th (12,6),(11,7),(10,8)

for 7th (10,11)

just count the possibility and dat is ur ans

12