Question

Consider the set $A=\{1,2,3,...,30\}$. The number of ways in which one can choose three distinct numbers from $A$ so that the product of the chosen numbers is divisible by 9 is

• A. 1590
• B. 1505
• C. 1110
• D. 1025

Answer 1

The correct option is A 1590

Given :
$A=\{1,2,3,...,30\}$

Numbers which are divisible by $3$ but are not by $9$:
$B=(3,6,12,15,21,24,30)$ ,
Total - $7$ numbers
Number divisible by $9$:
$C=(9,18,27)$,
Total - $3$ numbers
Rest of the numbers:
$D=(A-(B\cup C\left)\right)$,
Total - $20$ numbers.

Choosing 3 numbers whose product is divisible by 9 =
Total number of ways of choosing 3 numbers - Total number of ways of choosing 3 numbers from $D$ - One number is multiple of 3 but other two are choosen from set $D$.
Required permutation
$={(}^{30}{C}_3)-{(}^{20}{C}_3)-{(}^7{C}_1{\times }^{20}{C}_2)=\frac{30\times 29\times 28}{6}-\frac{20\times 19\times 18}{6}-7\times \frac{20\times 19}{2}=(5\times 29\times 28)-(20\times 19\times 3)-(7\times 10\times 19)=4060-2470=1590$