Question 632059: In how many distinct ways can the TORONTO be arranged
Found 2 solutions by ewatrrr, Theo:
Answer by ewatrrr(24785)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! 7!/(3!*2!) = 7*6*5*4*3*2*1 / 3*2*1*2*1 which is equal to 420
7! if all the letters were different.
Since 3 O's are common, you have to divide by 3!.
Since 2 T's are common, you have to divide by 2!.
Hard to see with 7 letters.
Easier to see with 3.
Assume letters are a,b,c
possible permutations are:
a,b,c
a,c,b
b,a,c
b,c,a
c,a,b
c,b,a
that's a total of 6 which is equal to 3!
Now assume 2 of the letters are the same.
let's assume it's b.
then the letters are a,b,b
possible permutations are:
a,b,b
b,a,b
b,b,a
that's a total of 3 which is equal to 6! / 2! = 3
same concept applies to 7 letters but the number of permutations is too large to list separately.
TORONTO has 2 of the letter T and 3 of the letter O and 1 each of the rest.
formula becomes 7! / (3!*2!) which is equal to 7*6*5*4*3*2*1 / 3*2*1*2*1 which is equal to 7*6*5*2 which is equal to 420.
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