Two distinct numbers are picked randomly from the first ten natural numbers. Find the probability that the L.C.M of the two numbers is 6.

Hint: Write the number of all possible ways in which two distinct numbers can be selected out of first ten natural numbers can be calculated using the formula, $^n{C_r}$, where $n$ is the total numbers and $r$ is equals to 2. Then find the total number of possible cases which gives L.C.M equals to 6. Then, find the required probability.

Complete step-by-step answer:
First of all, write the number of all possible ways in which two distinct numbers can be selected out of the first ten natural numbers can be calculated using the formula, $^n{C_r}$.
We use combinations to find total possible ways because order of the number does not matter.
We have to select 2 numbers from 10 numbers, hence, the number of possible ways will be,
$^{10}{C_2}$ which is equals to $\dfrac{{10!}}{{2!8!}} = \dfrac{{10.9.8!}}{{2.1\left( {8!} \right)}} = \dfrac{{90}}{2} = 45$
Therefore, the total number of outcomes is 45.
Now, we need to calculate the pairs that give L.C.M as 6.
We have $\left( {1,6} \right)$ and $\left( {2,3} \right)$ that will give the L.C.M as 6.
Therefore, the total number of favourable outcomes are 2.
Now, we will calculate the probability using the formula, $\dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of possible outcomes}}}}$
Hence, the required probability is $\dfrac{2}{{45}}$.

Note: We will use combinations to find total possible ways because order of the number does not matter .We just need to select the pairs, that is, whether we take 2,3 or 3,2 it means the same pair as it has the same numbers in the pair. The formula used in the calculation of probability is $\dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of possible outcomes}}}}$