# Question regarding the probability and combination

Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. Find the probability that the numbers in AP.

In the above problem, my teacher told that the total number of ways will be $21C2$. I understood this point. But there is one more problem given in the book:

Let two numbers are selected at random from set ${1,2,....,50,51}$, then find the probability that the sum of two numbers is even.

In this problem, the solution manual is taking the total number of ways of select two numbers as $51×51$.

I am not able to distinguish between these two problems and why aren't we using the combination in the second one too just like the first one. Any help would be appreciated.

Thanks.

• – user Mar 24 '18 at 8:08

In the first one we are selecting a pair among 21 object that is $\binom{21}{2}$ and can be viewed as a selection without replacement, that is $\frac{21\cdot20}2$.