This is a probability question.

Question

Ron finds 9 books at a bookstore that he would like to buy, but he can afford only 5 of them. In how many ways can he make his selection? How many ways can he make his selection if he decides that one of the books is a must?

Answer 1

At first, this seems like a tricky question, but it is actually quite simple. 

 

When the question says "One of the books is a must", you may simply subtract one of the books from both the 9 books at the bookstore as well as the 5 books he needs to purchase.

This is because the probability of him choosing this book is 100% or 1. So you can disregard it.

 

So the new question (disregarding the previous) is the following: 

 

Ron finds 8 books at a bookstore that he would like to buy, but he can afford only 4 of them. In how many ways can he make his selection?

 

This is a permutation function ( 8 choose 4 ). 

 

Pretend Ron chooses books at random and puts them in his cart while shopping. Let's go as far as pretending that Ron closes his eyes while he picks books, so he doesn't know which one he is picking. 

 

Let's pretend he can choose from the following:

 

Harry Potter

Hunger Games

1984

Animal Farm

Divergent

Chronicles of Narnia

Brave New World

Of Mice and Men

 

What are the chances of Ron choosing 1984 on the first pick?

1/8

 

What about the second pick?

1/7

(Remember there are only 7 now)

 

Repeat this two more time and you'll find you have these probabilities. 

 

1/8, 1/7, 1/6, 1/5

 

Multiply these together (we are assuming independence) and what do you get?

1/1680

 

Take the reciprocal of that number and you get the total possible combinations.. 1680.