Question

The number of ways in which one can select three distinct integers between 1 and 30 both inclusive whose sum is even is

Answer 1

Hint: When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination. Here, we are given that, number of integers is 1 to 30. So, numbers can be either all even or one even and other two odd. Thus, we need to select 3 distinct integers, so that their sum is even.
We will use the combination formula \[{}^n{C_r} = \dfrac{{(n!)}}{{(r!)(n - r)!}}\] also to find the final output.

Complete answer:
Given that,
Number of integers = 1 to 30
We need to select three distinct integers between 1 to 30, such that their sum is even.
Thus, numbers can be either all even or one even and other two odd.
Required number of ways is as
\[ = {}^{15}{C_3} + {}^{15}{C_1} \times {}^{15}{C_2}\]
Here, we will use the formula for a combination of choosing r unique ways from n possibilities as:
\[{}^n{C_r} = \dfrac{{(n!)}}{{(r!)(n - r)!}}\]
We will first find the values of each as below:
 First,
\[{}^{15}{C_3}\]
\[ = \dfrac{{(15!)}}{{(3!)((15 - 3)!)}}\]
On evaluating this, we will get,
\[ = \dfrac{{(15!)}}{{(3!)(12!)}}\]
Remove the factorial of denominator as below, we will get,
\[ = \dfrac{{(15 \times 14 \times 13 \times 12!)}}{{(3 \times 2 \times 1)(12!)}}\]
On simplifying this and removing the brackets, we will get,
\[ = 5 \times 7 \times 13\]
\[ = 455\]
Next,
\[{}^{15}{C_1}\]
\[ = \dfrac{{(15!)}}{{(1!)((15 - 1)!)}}\]
On evaluating this, we will get,
\[ = \dfrac{{(15!)}}{{(1!)(14!)}}\]
We know that, \[1! = 1\] and so applying this, we will get,
\[ = \dfrac{{(15 \times 14!)}}{{(1)(14!)}}\]
On simplifying this and removing the brackets, we will get,
\[ = 15\]
And,
\[{}^{15}{C_2}\]
\[ = \dfrac{{(15!)}}{{(2!)((15 - 2)!)}}\]
On evaluating this, we will get,
\[ = \dfrac{{(15!)}}{{(2!)(13!)}}\]
We know that, \[2! = 2 \times 1\] and so applying this, we will get,
\[ = \dfrac{{(15 \times 14 \times 13!)}}{{(2 \times 1)(13!)}}\]
On simplifying this and removing the brackets, we will get,
\[ = 15 \times 7\]
\[ = 105\]
Now, we will use all these values and substitute it as below:
\[{}^{15}{C_3} + {}^{15}{C_1} \times {}^{15}{C_2}\]
\[ = 455 + 15 \times 105\]
On evaluating this, we will get,
\[ = 455 + 1575\]
\[ = 2030\]
Hence, the number of ways in which one can select three distinct integers between 1 and 30 both inclusive whose sum is even is \[2030\] .

Note:
Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. And, permutation relates to the act of arranging all the members of a set into some sequence or order.