SOLUTION: In a certain lottery there are 49 numbers to choose from and 6 numbers can be chosen. In how many ways can the numbers be chosen? Please explain using 49!, ^! etc
Algebra ->
Permutations
-> SOLUTION: In a certain lottery there are 49 numbers to choose from and 6 numbers can be chosen. In how many ways can the numbers be chosen? Please explain using 49!, ^! etc
Log On
Question 355706: In a certain lottery there are 49 numbers to choose from and 6 numbers can be chosen. In how many ways can the numbers be chosen? Please explain using 49!, ^! etc Answer by jrfrunner(365) (Show Source):
You can put this solution on YOUR website! if order was important, then you would use permuations
49P6=49!/(49-6)!
=49!/43!
=49*48*47*46*45*44
--
typically in a lottery the "order" does not matter, so you would be using combinations not permutations.
a combination is basically a permutation where you back out same group of items that happen to be in different order
49C6=49P6/6!
=49!/(6!*(49-3)!)
=49!/(6!*43!)
=49*48*47*46*45*44/(6*5*4*3*2)