Question 1197177: How many arrangements of the letters in KILLNACOUNTING are possible?
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
There are 14 letters in "KILLNACOUNTING"
The letters that repeat are I, L, and N.
If we could distinguish between the repeated letters, then we'd have 14! = 14*13*12*11*10*9*8*7*6*5*4*3*2*1 = 87,178,291,200 different permutations.
The exclamation mark stands for a factorial. We start at 14 and count our way down to 1 multiplying along the way.
Many calculators have an exclamation mark button to make this calculation fairly quick. Meaning you can avoid typing in 14 times 13 times ... all the way down to 1.
Anyways, that very large number would be the answer if we could tell those repeated letters apart.
But we can't tell them apart.
There are...
2 I's
2 L's
3 N's
So we must divide that massive number by 2!*2!*3! = 2*2*6 = 24 to correct for the fact we cannot distinguish these letters from one another.
(87,178,291,200)/(24) = 3,632,428,800
This is approximately 3.63 billion
Answer: 3,632,428,800
Answer by ikleyn(52765)  (Show Source):
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