Question 273360
if ten digits 0 to 9 are arranged randomly with no repetition of digits how many different arrangements are possible?
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Take an example, say:  

              0842971536

There are 10 ways to pick a digit to go where the 0 is.
That leaves 9 ways to pick a digit to go where the 8 is.
That leaves 8 ways to pick a digit to go where the 4 is.
That leaves 7 ways to pick a digit to go where the 2 is.
That leaves 6 ways to pick a digit to go where the 9 is.
That leaves 5 ways to pick a digit to go where the 7 is.
That leaves 4 ways to pick a digit to go where the 1 is.
That leaves 3 ways to pick a digit to go where the 5 is.
That leaves 2 ways to pick a digit to go where the 3 is.
That leaves 1 way to pick a digit to go where the 6 is.

Answer 10*9*8*7*6*5*4*3*2*1 = 10!= 3628800

Edwin</pre>