Permutations and Combinations 4

[JordanUpiter]

Ok I am a little confused. Would it work if i wrote. (1/60)*(1/59)*(1/58)*(1/57) Since u have a 1 in 60 chance to get the first number then 1 in 59 chance for second and so on would this work? but i dont get the same answer

[azeenator]

Khan you just saved my life! :)

[Dragonnoodle1]

So basicly if the order was important you would use permutations?

[PaidVideoReviews]

@Dragonnoodle1 Yes.

[Dragonnoodle1]

I have a test of this in an hour and I didnt really study o.O I am going to die!

[l4l01234]

Excellent video.

[Ekman900509]

These 4 vids learned me everything about permutation and combinations that I tried to learn with my calculus book for hours.

[moka22051]

@Ekman900509 YYYESS!!

[PhxPride1]

Watched the four video set and loved it, thanks Khan.

[mitomke]

So each lottery is an independent event with a one in 487,635 chance in winning. is this correct? So if the same 4 numbers are played week after week , the chance of winning with those same numbers is not increased because each event is independent.

[24ballin23]

His answer is wrong. Its [24/ 487635] because the order doesn't matter thus those winning numbers occur in 4!=24 ways.

[crystalidx]

@24ballin23 He did do that, because order doesn't matter, he divided it by 4! and he got 487635 for the combinations. And for the possibity he got 1/487635. its not wrong.

[treefangers]

@24ballin23 They occur in 24 different ways; however, there are 11,703,240 permutations - not 487,635.

24/11703240 = 1/487635

Tacos are delicious.

[shdubbleorti]

how would you solve one an outrageously high number such as. 300P289?

[Waranle]

Thank you Sal, plz more videos of discrete maths :)

[MomoBG22]

the old calc was way cooler :))