Stupid question

Suppose you arrange ten given names randomly, what are the odds they'll end up in alphabetical order?

1 : 3628800.

Think about the number of ways to write a 2 digit number. There are 2: 12 and 21.

Think about the number of ways to write a 3 digit number. There are 6: 123, 132, 213, 231, 312, 321.

Think about the ways to arrange a four digit number, 1-2-3-4. There are twenty four combinations available to write the number (1234), but only one way to write 1-2-3-4. You can try it for yourself... 1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321.

The math formula for the number of combinations is simply n! (n factorial) where n is the number of names or numbers. Since n = 10, 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.

Similarly for a name, the odds of getting in it the exact same order is 1:(10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 1:3,628,800.

Thanks.

BTW that was so easy I should have figured it out myself. I'm bad at math, but I did take an elementary stats and probability course in the not impossibly distant past, and I knew that one!

Thanks again.

=FACT(10)=3628800
=PERMUT(10,10)=3628800 are the formulas in excel.

=PERMUT(10,3)=720 (=10*9*8) because you are choosing three things out of ten and the order matters.


=COMBIN(10,3)=120 (=FACT(10)/(FACT(7)*FACT(3)) ) is choosing 3 things out of 10 but order does not matter.

FACT -> FACTORIAL
PERMUT -> PERMUTATION
COMBIN -> COMBINATORIAL