In a lottery game, a player picks 7 numbers from 1 to 47. How many different choices does the player have if order doesn't matter?
2 answers
47C7 = 62891499
You are choosing 7 from 47, that is called
C(47,7)
= 47!/(7! 40!)
My calculator has the nCr button and gave me 62,891,499
if not, then
47!/(7! 40!)
= 47*46*45*44*43*42*41*40*39*...*1 / ((40*39*....1)(7*6*5*4*3*2*1))
= 47*46*45*44*43*42*41 / (7*6*5*4*3*2*1)
= 62,891,499
C(47,7)
= 47!/(7! 40!)
My calculator has the nCr button and gave me 62,891,499
if not, then
47!/(7! 40!)
= 47*46*45*44*43*42*41*40*39*...*1 / ((40*39*....1)(7*6*5*4*3*2*1))
= 47*46*45*44*43*42*41 / (7*6*5*4*3*2*1)
= 62,891,499
2 answers