Any natural number is divisible by both 2 and 3 if it is divisible by (2×3=) 6
So, the numbers divisible by 6 from 1 to 100 are 6,12,18,....96
96=6+(n-1)6 {where 'n' is no.of terms }
On solving, we get n=16
So, there are total 16 numbers out of which the 3
distinct numbers are to be chosen.
Total no. of favourable choices =16×15×14=n(A)
Total no.of ways=100×99×98=n(S)
P(divisible by 2 and 3)=n(A)/n(S)=(100×99×98)/(16×15×14)=4/1155 (answer)
