Question

How many ways are there to choose three different numbers each between one and a
hundred so that their sum is even? Explain

Answer 1

In between 1 to 100 there are 98 numbers. Out of them 49 are even and 49 are odd.

First we're considering that the orientation of numbers being picked doesn't matter. If it matters then, we'll have to multiply the result with the no of permutations we can get out of three numbers=3P3=6.Here P means permutation.

Now, addition of three numbers make an even number if:

All the three numbers are even.For this option, we will have (49C3) ways. Here C means combination.

Two of them are odd and the remaining number is even. For this option, we will have (49C2)*(49C1) ways.

So, the total no. of ways:

(49C3)+(49C2)(49C1)

=(49!/46!3!)+(49!/47!2!)(49!/48!1!)

=18424 +49*1176

=76048

And, if the orientation of numbers matter then the no of ways will be 76048*6=456288

So, we'll have two answers depending on the orientation of numbers:

If orientation doesn’t matter and all of them are considered as same then, the no of ways = 76048

If orientation matters then, the no of ways = 456288