Question

Let $n_1$ and $n_2$ represent respectively the number of possible ordered and unordered triplets $(a,b,c)$ such that $abc=144,(a,b,c\in N)$, then

• A. $n_1=90$
• B. $n_1=45$
• C. $n_2=15$
• D. $n_2=18$

Answer 1

Answer: (A), (D)

$n_2=18$

$abc=90=2^4\times 3^2$
Let $a=2^{p_1}\times 3^{q_1}b=2^{p_2}\times 3^{q_2}c=2^{p_3}\times 3^{q_3}\Rightarrow 2^{p_1+p_2+p_3}\times 3^{q_1+q_2+q_3}=2^4\times 3^2\Rightarrow p_1+p_2+p_3=4,q_1+q_2+q_3=2\text{number of non-negative solutions of the above equations}=n_1{=}^6{C}_2{\times }^4{C}_2=90$

For $n_2$,
The pairs with two numbers equal is $(1,1,144),(2,2,36),(3,3,16),(4,4,9),(6,6,4),(12,12,1),$ all other pairs are different.

$n_2=\frac{90-18}{6!}+6=18$