Question

The number of ordered triplets of natural numbers $(a,b,c)$ for which $abc\le 11$ is

• A. $52$
• B. $53$
• C. $55$
• D. $56$

Answer 1

The correct option is D $56$

$abc=1$ in $1$ way

Consider $abc=2$
This can be done by the permutation of $1,1,2$ which can be done in $\frac{3!}{2!}=3$ ways
So, $abc=2,3,5,7,11$ in $3\times 5=15$ ways
Similarly, $abc=4,9$ in $6\times 2=12$ ways
$abc=6,10$ in $9\times 2=18$ ways
$abc=8$ in $1+6+3=10$ ways
So, total number $=56$