First I wrote out all the possibilities for EMMA using the letters A, B and C instead of E, M and A.
These are my results:
ABBC
ABCB
ACBB
BABC
BACB
BBCA
BBAC
BCAB
BCBA
CABB
CBAB
CBBA
There are 12 different arrangements.
A shorter way of finding this is by using the number of arrangements for a four-letter word with all different letters.
In my table of results I found that by using the equation n(n-1) x n(n-2) ... 3x2x1 equals the number of arrangements for any number of letters. Then I looked up an A-Level text book which explained that n(n-1) x n(n-2) ... 3x2x1 is equal to n factorial (n!). Using this I experimented with the different sums to try and work out a direct way to find out the number of different arrangements for a word which has 2 letters the same.
These are my results:
ABBC
ABCB
ACBB
BABC
BACB
BBCA
BBAC
BCAB
BCBA
CABB
CBAB
CBBA
There are 12 different arrangements.
A shorter way of finding this is by using the number of arrangements for a four-letter word with all different letters.
In my table of results I found that by using the equation n(n-1) x n(n-2) ... 3x2x1 equals the number of arrangements for any number of letters. Then I looked up an A-Level text book which explained that n(n-1) x n(n-2) ... 3x2x1 is equal to n factorial (n!). Using this I experimented with the different sums to try and work out a direct way to find out the number of different arrangements for a word which has 2 letters the same.
