If you do not divide the amount of repeated letters by the total combinations with different letters you will be writing down some same combinations.
1 × 2 × 3 × 4 ÷ 1 ÷ 2 = 12 combinations for the word EMMA
I shall be showing this works by using different words.
Different arrangements for the word JAN (three letter word)
Number of combinations= 6
1 × 2 × 3 = 12 combinations for the word JAN
Different arrangements for the word BOB (three letters with same letters)
Number of combinations= 3
1 × 2 × 3 ÷ 1 ÷ 2 = 3 combinations for the BOB
Different arrangements for the word YU (two letter word)
Number of combinations= 2
1 × 2 = 2 combinations for the word SU
Different arrangements for the word SS (two letter word with same letters
Number of combinations= 1
1 × 2 ÷ 1 ÷ 2 = 1 combination for the word SS
Formula meanings
n = number of letters in the word
y = number of repeated letters
! = factorial
My formula
My formula for word with different letters is
n! =
n × (n – 1) × (n – 2) . . . 2 × 1
This is because it is quicker to find the total combinations for a certain amount of letters without writing them all up.
The formula for letters that have been repeated in the word is
n! ÷ y!=
n × (n – 1) × (n – 2) . . . 2 × 1
y × (y – 1) × (y – 2) . . . 2 × 1
This is because without dividing you will get some same combinations when written up.
Total combinations for words with letters up to ten
Total combinations for words with repeated letters up to ten
Z= total number of combinations with same letters
Extension
What I am going to do now is find a formula for a word with different sets of same letters.
Different arrangements for the DEDE (four letter word with sets of same letters)
Number of combinations= 6
1 × 2 × 3 × 4 ÷ 1 ÷ 2 ÷ 1 ÷ 2 = 6 combinations for the word DEDE
Different arrangements for these letters DDDEE (five letters all together with sets of same letters)
Number of combinations= 10
1 × 2 × 3 × 4 × 5 ÷ 1 ÷ 2 ÷ 3 ÷ 1 ÷ 2 = 10 combinations for the letters DDDEE
The formula for this section is
n! ÷ y! ÷ y! . . . y! ÷ y! =
n × (n – 1) × (n – 2) . . . 2 × 1
y × (y – 1) × (y – 2) . . . 2 × 1 y × y . . . y
This is because it will make it easier with things with more letters and you may also write down some same combinations.
This is a table showing more set of letters in a certain amount of letters up to ten
Conclusion
In this coursework I have discovered that the letters in a word or a lot of letters put together can be re-changed to find different combinations. For with and without same letters I have a formula to make it easier to the total of combinations without the working out. I have also discovered it would work other situation e.g. when there is more sets of same letters.
