Answer:
She can chose 2 books from 8 books in 28 different way.
Explanation:
In the given scenario the member has to chose 2 books from a group of 8 books. The order of choosing books does not matter. For example if the books are A,B,C,D,E,F,G and H, and the member picks books A and B or books B and A, she is picking up the same books. This means order of selection does not matter here, so this is a problem of combinations.
We have to form combination of 8 books taken 2 at a time i.e. 8C2
The general formula of combination is:
![nCr=(n!)/(r!(n-r)!)](https://img.qammunity.org/2020/formulas/mathematics/high-school/nwmlafk3j9ekq8l7shaqupgq9ep02649pd.png)
Using the values of n=8 and r=2, we get:
![8C2=(8!)/(2!(8-2)!)=28](https://img.qammunity.org/2020/formulas/mathematics/middle-school/d9uwdedyxwch1p6oq64vvq3f2bbwa8ffyp.png)
This means, she can chose 2 books from 8 books in 28 different ways.