Question 879135: How many distinguishable ways can be written using all the letters in the word ALGEBRA?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! Number of letters in this word are:
A = 2
L = 1
G = 1
E = 1
B = 1
R = 1
There are 7 total letters with 2 of them being the same.
The formula is 7! / 2! = 2520.
This is hard to show because there are so many possibilities, but I can show it with a simpler type problem.
Assume the letters ABC.
The number of possible arrangements are 3! = 6
Those arrangements are:
ABC
ACB
BAC
BCA
CAB
CBA
Now assume the letters AAC.
The number of possible arrangements are 3! / 2! = 3.
Those arrangements are:
AAC
ACA
CAA
You have essentially replace the B with another A.
Where you had ABC and ACB, you now only have AAC and ACA
Where you had BAC and BCA, you now only have AAC and ACA
Where you had CAB and CBA, you now only have CAA and CAA.
You now have AAC twice and ACA twice and CAA twice.
The number of unique arrangements is only 3 which are AAC and ACA and CAA.
The general formula for number of unique permutations is:
P = n! / (x1! * x2! * ... xn!) where:
n is the number of letters.
x1, x2, ..., xn are the number of letters that are the same.
For example:
Original letters are ABCDEFG
Formula is 7!
Original letters are AACCEEG
Formula is 7! / (2! * 2! * 2!)
Original letters are AAAAABB
Formula is 7! / (5! * 2!)
That's how it works.
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