SOLUTION: How many ways are there to select 3 books from a shelf of 20 books?

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Question 284028: How many ways are there to select 3 books from a shelf of 20 books?
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If order matters, then there are 20 ways to select the first book, 19 to select the second, and 18 to select the third (since we're not replacing the books). Multiply these values to get 20*19*18 = 6840


So there are 6840 different ways to select 3 books from a shelf of 20 where order matters.



If order does NOT matter, then you will have duplicates to worry about. It turns out that each triple (ie each selection of 3) occurs 6 times. For instance, if you have books 'a', 'b', and 'c', then you will have the following:

abc
acb
bac
bca
cab
cba


Notice there are 6 ways to order 'a', 'b', and 'c'


If order does NOT matter, then you must divide the previous result of 6840 by 6 to get 1140. Dividing by 6 will eliminate the counting of duplicates. So there are 1140 different ways to select 3 books from 20 where order does NOT matter.


Formally, we just computed the permutation of 20 choose 3 for the first answer and a combination of 20 choose 3 for the second.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


It actually depends on whether or not the order in which you select the books matters. One must presume that the shelf contains 20 different books as well because that fact is a factor in answer to this question.

If you don't care about the order, you need the number of combinations of 20 things taken 3 at a time:



On the other hand, if the subset War and Peace, Differential Equations for Dummies, and The Cat in the Hat, is different from the subset The Cat in the Hat, War and Peace, and Differential Equations for Dummies, then you need the number of permutations of 20 things taken 3 at a time:



Notice that the only difference is the 3! that is in the denominator of the combination calculation but not in the permutation calculation.

You can do your own arithmetic.

John