math
posted by Em .
annie,betty,Cathy and Danial are going to stand on a row for taking pictures. How many ways can they stand?

4*3*2*1

The first person has the cohoice of 4 spots, takes one
The second person has choice of 3 spots, takes one
The third person has choice of two spots takes one
The fourth person is stuck with the last spot, takes it
4*3*2*1 = 24
permutations of n items taken one at a time. 
is n !
n (n1) (n2) ....... 1 
By the way the permutations of n items taken r at a time is
n!/(nr)!
and if it does not matter how the r items are arranged in each sub group, then it is the number of "combinations"
n! / [ r! (nr)! ] 
Now what if they were to be paired and we care if each is on left or right in the photo?
that is permutations of four people taken 2 at a time
4!/2! = 4*3*2*1/(2*1) = 12 ways
But what if we do not care if they are on the right or the left. (BettyCathy the same as CathyBetty)
4!/[2!(42)!] = 4*3/2 = 6 ways 
Well, that was fun, have not done that for a while.

:)

LOL  well there is much more but it is probably not on Em's list for tonight (Pascal's triangle, binomial theorem). Do not get me started :)
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