Question 924181
Re TY
If ALL ten letters were different: 10! would be the number of ways of arranging them
For ex: 5 different books on a shelf: 5! = 120, would be the ways of arranging them
......
When there are duplicates like the following, it cuts down the number of ways...

P R K T S V R J P F  (10Letters: 2Ps, 2Rs)
10!/(2!2!) = ways of arranging = 907,200
10! = {{{10*9*8*7*6*5*2*1}}} = 3,628,800
Nice to have a Calculator with a ! button