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Question: The number of ways in which one can select three distinct integers between 1 and 30 both inclusive w...

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

Explanation

Solution

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 nCr=(n!)(r!)(n−r)!{}^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
=15C3+15C1×15C2= {}^{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:
nCr=(n!)(r!)(n−r)!{}^n{C_r} = \dfrac{{(n!)}}{{(r!)(n - r)!}}
We will first find the values of each as below:
First,
15C3{}^{15}{C_3}
=(15!)(3!)((15−3)!)= \dfrac{{(15!)}}{{(3!)((15 - 3)!)}}
On evaluating this, we will get,
=(15!)(3!)(12!)= \dfrac{{(15!)}}{{(3!)(12!)}}
Remove the factorial of denominator as below, we will get,
=(15×14×13×12!)(3×2×1)(12!)= \dfrac{{(15 \times 14 \times 13 \times 12!)}}{{(3 \times 2 \times 1)(12!)}}
On simplifying this and removing the brackets, we will get,
=5×7×13= 5 \times 7 \times 13
=455= 455
Next,
15C1{}^{15}{C_1}
=(15!)(1!)((15−1)!)= \dfrac{{(15!)}}{{(1!)((15 - 1)!)}}
On evaluating this, we will get,
=(15!)(1!)(14!)= \dfrac{{(15!)}}{{(1!)(14!)}}
We know that, 1!=11! = 1 and so applying this, we will get,
=(15×14!)(1)(14!)= \dfrac{{(15 \times 14!)}}{{(1)(14!)}}
On simplifying this and removing the brackets, we will get,
=15= 15
And,
15C2{}^{15}{C_2}
=(15!)(2!)((15−2)!)= \dfrac{{(15!)}}{{(2!)((15 - 2)!)}}
On evaluating this, we will get,
=(15!)(2!)(13!)= \dfrac{{(15!)}}{{(2!)(13!)}}
We know that, 2!=2×12! = 2 \times 1 and so applying this, we will get,
=(15×14×13!)(2×1)(13!)= \dfrac{{(15 \times 14 \times 13!)}}{{(2 \times 1)(13!)}}
On simplifying this and removing the brackets, we will get,
=15×7= 15 \times 7
=105= 105
Now, we will use all these values and substitute it as below:
15C3+15C1×15C2{}^{15}{C_3} + {}^{15}{C_1} \times {}^{15}{C_2}
=455+15×105= 455 + 15 \times 105
On evaluating this, we will get,
=455+1575= 455 + 1575
=2030= 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 20302030 .

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.