Question

How many different combinations of product can be obtained by multiplying three numbers from $12,13,14,15,16,19$ without using same number twice?

• A. $20$
• B. $\frac{6!}{3!}$
• C. $6!$
• D. $120$

Answer 1

The correct option is A $20$

We are given with $6$ different numbers out of which we have to select $3$ numbers at a time and find number of all possible combinations of selecting $3$ numbers.

We also know that number of ways of selecting "r" different things out of "n" different things is equal to ${}^n{C}_r=\frac{n!}{(n-r)!.r!}$

According to question $n=6$ and $r=3$

So, Number of combinations is $={}^6{C}_3$
$\Rightarrow {}^6{C}_3=\frac{6!}{(6-3)!\times 3!}=\frac{6!}{3!.3!}$

$\Rightarrow {}^6{C}_3=\frac{720}{36}=20$