# In how many ways can we choose 8 books out of 10?

txmedteach | Certified Educator

If order does not matter, you end up performing a different calculation than the one above. The one above treats the following choices separately (let's label the books a-j):

1)a,b,c,d,e,f,g,h

2)a,b,c,d,e,f,h,g

Notice that they contain the same 8 books, but the only difference is that their order is different.

If you want to take out the effect of counting different orders, you need to determine for each "set" of books you pick out, how many ways you can order them. You then divide by the number of ways they can be ordered in order to determine how many unique combinations there are.

In our case, we are picking out sets of 8, so we'll consider how many ways we can order a set of 8 books. Let's treat it like we have 8 sequenced spaces. In space 1, we have 8 choices for books. In space 2, we have the remaining 7 choices. In space 3, we have the remaining 6 choices, and so on. This can be represented mathematically by:

Total ways to order = `8*7*6*5*4*3*2*1 = 8! = 40320`

So, there are 40,320 ways to order each unique set of books. So, what we can do to find the number of unique sets is to take the number we got from sciencesolve through his calculation of the permutations and divide it by 40,320:

Number of combinations = `1814400/40320 = 45`

So, we find that there are actually only 45 unique ways to select 8 books from 10!

Side Note: This is the same answer you would get if you were to figure out how many ways you could choose 2 books from the 10 available. The way to think about this is that finding how many ways you can choose 8 books from the 10 available is equivalent to finding how many ways you could reject 2 books from the available 10 (which is, in effect choosing 2 books out of the 10). I always found that interesting. Fun Fact, I suppose.

sciencesolve | Certified Educator

You need to use arrangements to find the number of ways you may choose 8 books out of 10.

You need to use factorial formula such that:

`P_n^k = (n!)/((n-k)!)`

Substituting 10 for n and 8 for k yields:

`P_10^8 = (10!)/((10-8)!) =gtP_10^8 =(10!)/(2!)`

`P_10^8= (2!*3*4*...*9*10)/(2!)`

`P_10^8 = 1814400`

Hence, you may choose 8 books out of 10 in 1814400 ways.