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can any1 help me with the justification of the emmas dilemma formula?

the formula is:

X!

---- = number of arrangerments

A!B!C! etc.

A/B/C being the the number of repeats

X being number of letters

e.g

aaabb

formula=

5!

----

3!2! (due to 3 a's and 2 b's)

can any1 help me justify and prove this?

the formula is:

X!

---- = number of arrangerments

A!B!C! etc.

A/B/C being the the number of repeats

X being number of letters

e.g

aaabb

formula=

5!

----

3!2! (due to 3 a's and 2 b's)

can any1 help me justify and prove this?

well look at it this way:

You have X letters to pick

You pick them one by one, and lay them out in the order you pick them

For the first letter there are X possiblities

For the second letter there are X-1 possiblities

and so on, until you have X! different ways of arranging the letters

BUT THEN you notice that some of those are the same - they arent all unique ways of ordering them. Lets say that two of the letters available to choose were the same. That means that for every arrangement of letters, there would be one other identical arrangement - because you could just swap the two matching letters.

OK so what if there was 3 of the same letters. Well then you will have every unique arrangement repeating 6 times, since there are six different ways of order the 3 repeated letters.

So you do X! then divide it by A! where A is the number of duplicated letters

Then the result you get, you can imagine checking it again, and seeing if you have any duplications in the arrangements. If you had another few identical letters available then once again you would have duplications, the number of duplications of each arrangement equaling the number of ways of arranging those letters - so you divide again by B!, and so on

You have X letters to pick

You pick them one by one, and lay them out in the order you pick them

For the first letter there are X possiblities

For the second letter there are X-1 possiblities

and so on, until you have X! different ways of arranging the letters

BUT THEN you notice that some of those are the same - they arent all unique ways of ordering them. Lets say that two of the letters available to choose were the same. That means that for every arrangement of letters, there would be one other identical arrangement - because you could just swap the two matching letters.

OK so what if there was 3 of the same letters. Well then you will have every unique arrangement repeating 6 times, since there are six different ways of order the 3 repeated letters.

So you do X! then divide it by A! where A is the number of duplicated letters

Then the result you get, you can imagine checking it again, and seeing if you have any duplications in the arrangements. If you had another few identical letters available then once again you would have duplications, the number of duplications of each arrangement equaling the number of ways of arranging those letters - so you divide again by B!, and so on

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