Q:

Choosing $3$ numbers from set $\{ 1,2, \dots, 60 \}$ such that product of them is a multiple of $13$

My try
There's only $4$ numbers from given set, that are multiple of $13$: $A = \{ 13,26,39,52 \}$
The condition is $d_1d_2d_3 (mod 13)=0$, so I should choose atleast one of these numbers from set $A$
if I choose $d_1$, from $A$, and $d_2,d_3$ from other numbers, then I have $\binom{4}{1} * \binom{60-4}{2}$ numbers
if I choose $d_1,d_2$ from $A$,and $d_3$ from others, then I have $\binom{4}{2} * \binom{60-4}{1}$ numbers
if I choose $d_1,d_2,d_3$ from $A$, then I have $\binom{4}{3}$ numbers
So in total I have $\binom{4}{1} \binom{56}{2} + \binom{4}{2} \binom{56}{1} + \binom{4}{3} = 6500$ numbers.
Is my approach true, or I have missed something?

A:

What you have is correct.  Alternatively you could compute the number of ways to choose 3 numbers from the set, and subtract off those choices which do not have one of the numbers which 13 divides.  In this case you have 56 to choose from, choose 3.  This way the number of choices would be $\binom{60}{3}-\binom{56}{3}$.

