© How many numbers, of 9 digit numbers, which have all different digits ?

Question

mary.rojas · Answer

so it can not repeat right? since it says "different".

goformit100 · Answer

ok

anonymous · Answer

There is 10 choices for the ones digit, 9 for the tens and going all the way to the 9th digit and so on. But note that the 9th digit on the left will only have 1 choice since it can't be 0 hence we get:$$\bf \frac{ 10! }{ 2 }=1814400 \ ways$$

mary.rojas · Answer

so remember that when you multiply you delete one and another each time.....like 
9*8*7*6*5*4*3*2*1 so you do not repeat any of the numbers... does that make sense?

anonymous · Answer

@goformit100 This is basically a permutations question so treat it that way..

goformit100 · Answer

Thank you @genius12 and @mary.rojas

goformit100 · Answer

|dw:1375986788851:dw|

anonymous · Answer

@goformit100 it's not 9!....

goformit100 · Answer

ok...

anonymous · Answer

@Luis_Rivera No you don't. There is 10 options for the first place value, 9 for the second, all the way to the last place value which only has 1 option since it can't be 0. Hence we get:$$\bf \frac{ 10! }{ 2 }=1,814,440 \ ways$$

goformit100 · Answer

ok I got it @genius12 Thanks

anonymous · Answer

yw