No. of Letters + 2 x No. of Letters = No. of Combinations
[ 3 + 2 x 3 = 15 ]
As shown above, it is clear that the method is incorrect as the answer we come to is wrong. Now I will use the factorial method.
No. of Letters Factorial = No. of Combinations
[ 3 ! = 1 x 2 x 3 = 6 ]
As shown above, the factorial method is correct as we come to the right answer. Hence, it is possible to finds the combinations of names with all different letters using the factorial method. The factorial method, gives us the correct number of combinations every time. The following table shows this.
Part 2 – The different combinations of Emma’s name
Below are all the different combinations of the name Emma.
- EMMA
- EAMM
- EMAM
- MEAM
- MAEM
- MMEA
- MMAE
- MEMA
- MAME
- AMME
- AEMM
- AMEM
As you can see, the name Emma has only 12 different combinations in contrast to the 24 the name Lucy has. This is because, Emma has two letters in it which are the same, so the number of combinations for a four (different) letter word, would be halved, to give the results.
Calculations _
We can use the following method to finds the different number of combinations for a four letter word with two letters the same.
No. of Letters Factorial ÷ No. of letters the same = No. of Combinations 4 ! ÷ 2 ! = 12
The above method is correct as it gives the correct number of combinations. This is because an arrangement of 4 different letters provides 24 different combinations, however in this case, 2 of the letters are the same, and as the combinations cannot be repeated, the number of letters factorial must be divided, by the number of repeated letters factorial. I can guarantee the above method is correct by taking the arrangement of letters, ABBBC. AS you can see this arrangement has 3 x B, 1 x A and 1 x C.
Below are all the different combinations for the arrangement of ABBBC.
- ABBBC
- ABBCB
- ABCBB
- ACBBB
- BABBC
- BABCB
- BACBB
- BCABB
- BCBAB
- BCBBA
- BBCBA
- BBCAB
- BBABC
- BBACB
- BBBCA
- BBBAC
- CABBB
- CBABB
- CBBAB
- CBBBA
Each repeated letter can be replaced by the other repeated letters. For example ABBBC can use three letters, rearranged, that still provide the same combination three times, therefore cutting the number of combinations significantly.
A B B B C
The arrangement ABBBC provides 20 different combinations. If you take number of letters in the arrangement factorial, 5!, you are provided with 120 combinations, but if you divide that by the number of repeated letters factorial, 3!, you are provided with only 20 combinations, which as you can see from above, is the correct number of combinations.
Therefore, for any number of letters, that has only one set of repeated letters, the formula below can be guaranteed to find the number of combinations.
No. of Letters Factorial ÷ No. of letters the same = No. of Combinations
Part 3 – Different combinations of various groups of letters.
If you take the arrangement of AABBB, you can find the number of combinations, through a slightly different formula, than the one used, to find the number of combinations of Emma.
Below are all the different combinations of AABBB.
- AABBB
- ABABB
- ABBAB
- ABBBA
- BABBA
- BBABA
- BBBAA
- BAABB
- BABAB
- BBAAB
As you can see from the list above, there are only 10 combinations for the arrangement of letters, AABBB. The formula below, also guarantees the number of combinations.
Number of Combinations =
Number of letters !
(Number of repeated letters1 ! x Number of repeated letters2 ! )
5 !
(2 ! X 3 !) = 10 combinations
The following table, shows some different combinations.
As you can see from the previous table for 2 or more sets of repeated letters. This means that as combinations cannot be repeated, you would need to multiply the factorials of the numbers of repeated letters, before you could use the following formula to calculate the correct number of combinations, X ! ÷ (n1 x n2 x etc) [X being the number of letters, and n being the number of repeated letters.].
Linking Formulas
The General Formula is No. of Combinations = X ! ÷ (n1 ! x n2 ! x etc.)
This formula can be applied to any set of letters as the formula works on the basis of primarily finding out the number of combinations for multiple sets of repeated letters. The n1 or n2 can be replaced with 1 in order to find out the number of combinations for one repeated set of letters. This is because there is no extra set of repeated letters, so X! eventually is divided by n!. You can replace the n1 and n2 by 1 in order to find out the number of combinations for a set of completely different letters. This is because if n is equal to 1 then it is the same as having no repeated letters so the formula can be simplified to X!.
There is no exception, as the formula allows you to work out the number of combinations for completely different letters, a set of repeated letters and multiple sets of repeated letters.
