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 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 $1590$

We define the following subsets of set $A:$

$X=\text{{9,18,27}}$ (divisible by $9$) Number of elements$=3$

$Y=\text{{3,6,12,15,21,24,30}}$ (divisible by $3$ but not by $9$) Number of elements $=7$

$Z=\text{all the remaining elements of set A}$ (not divisible by $3$) Number of elements $=20$

Number of ways of choosing $3$ distinct numbers such that their product is $\text{NOT}$ divisible by $9:$

$\text{(I)}$ Choose any $3$ numbers from subset $Z={}^{20}{\text{C}}_3=1140$

$\text{(II)}$ Choose any $2$ numbers from subset $Z$ and any $1$ number from subset $Y={}^{20}{\text{C}}_2\times {}^7{\text{C}}_1=1330$

Total number of ways of choosing $3$ distinct numbers such that their product is $\text{NOT}$ divisible by $9=1140+1330=2470$

Total number of ways of choosing $3$ distinct numbers from set $A={}^{30}{\text{C}}_3=4060$

So, Number of ways of choosing $3$ distinct numbers such that their product is divisible by $9=4060-2470=1590$