Question

A person going to a party was asked to bring 3 different bags of chips. Going to the store, she finds 13 varieties.How many different selections can she make?

Answer 1

In this problem we are given that a person who wants to go to a party would like to bring 3 different bags of chips; and also, that he/she has 13 varieties.

As she has differents options to choose, we see that this one is a problem of permutations or combinations. But the order of the chips the person brings does not matter, and so, we have to find a combination. We remember then that the equation for finding a combination of n objects and k possibilites is given by the following binomial coefficient

`\binom{n}{k}=(n!)/(k!(n-k)!)`

Where the symbol ! represents factorial (the product of the number to the number). For example,

`4!=4\cdot3\cdot2\cdot1=24`

Now, in our exercise, we have that the person has 13 varieties and has to choose 3 of them. We have to calculate then:

`\binom{13}{3}`

For doing so, we will start by replacing the values on the formula:

`\begin{gathered} \binom{13}{3}=(13!)/(3!(13-3)!) \\ =(13!)/(3!(10)!) \\ =(13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)/(3\cdot2\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) \end{gathered}`

And cancelling the equal terms, we obtain:

`\begin{gathered} =\frac{13\cdot12\cdot11\cdot\cancel{10}\cdot\cancel{9}\cdot\cancel{8}\cdot\cancel{7}\cdot\cancel{6}\cdot\cancel{5}\cdot\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}{3\cdot2\cdot\cancel{10}\cdot\cancel{9}\cdot\cancel{8}\cdot\cancel{7}\cdot\cancel{6}\cdot\cancel{5}\cdot\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}} \\ =(13\cdot12\cdot11)/(3\cdot2) \\ =(13\cdot12\cdot11)/(6) \\ =13\cdot2\cdot11=286 \end{gathered}`

This means that the person would be able to do 286 different selections.