SOLUTION: Mimi has eight books from the Statistics is Fun series. She plans on bringing two of the eight books with her in a road trip. (Show all work. Just the answer, without supporting wo
Algebra ->
Permutations
-> SOLUTION: Mimi has eight books from the Statistics is Fun series. She plans on bringing two of the eight books with her in a road trip. (Show all work. Just the answer, without supporting wo
Log On
Question 1132180: Mimi has eight books from the Statistics is Fun series. She plans on bringing two of the eight books with her in a road trip. (Show all work. Just the answer, without supporting work, will receive no credit).
(a) Does the order matter in the book selection?
(b) Based on your answer to part (a), should you use permutation or combination to find the number of the different ways the two books can be selected?
(c) How many different ways can the two books be selected? Found 2 solutions by addingup, ikleyn:Answer by addingup(3677) (Show Source):
You can put this solution on YOUR website! a) order doesn't matter
b) Combination
c) 28 ways
.
Pick 2 out of 8, order doesn't matter:
8C2 = nCr = n!/[(r)!(n-r)! = 8!/[(2)!(8-2)! = 40320/1440 = 28 ways
.
.
NOTE: permutation is when order matters. So, you know when we say "it's a combination lock"? Well, statistically it's a permutation lock because you have to enter the numbers in a specific sequence.
