N=1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21…., 999
Divisible by either 3 or 5 [ P(3 or 5) ] = Divisible by 3 [ P(3) ] + Divisible by 5[ P(5) ] – Divisible by both 3 & 5 [ P(3 & 5) ]
P(3) : 3, 9, 15,21,27,….,999
AP series : 999 = 3 + [n-1]6 ; n= 167
P(5) : 5,15,25,35,…,995
995 = 5 + [n-1]10 ; n = 100
P(3&5) : 15,45,75,105,135,..,990
990 = 15 + [n-1]30; n= 33.5 => 33
So, P(3 or 5) = 167 + 100 – 33 = 234