E.g. 5! = 5 x 4 x 3 x 2 x 1 5! = 120.
Factorial notation is symbolized using an exclamation mark!
I realized this is because if I could find the total number of combinations by multiplying the total number of letters by the previous number of combinations it was the same as multiplying the total number letters by its previous consecutive numbers Factorial Notation.
The formula to find out the total number of combinations for any name when all the letters are the same is:
Number of combinations = Total number of letters Factorial
tn = n!
This is explained in more detail in section 3a.
Part 3:
(a). Investigating the number of different arrangements of letters in names with varying numbers of letters, where all the letters are different:
Investigation:
Name: “D”
Number of letters: 1
Arrangements:
Total Number of Arrangements: 1
Name: “KO”
Number of letters: 2
Arrangements:
Total Number of Arrangements: 2
Name: “TIM”
Number of letters: 3
Arrangements:
Total Number of Arrangements: 6
Name: “LUCY”
Number of letters: 4
Arrangements:
Total Number of Arrangements: 24
Name: “FRANK”
Number of letters: 5
Arrangements:
Total number of arrangements: 120
Summary of Part 3a:
We can see a pattern between the number of arrangements and the number of letters.
Where n = number of letters,
1 2 3 4 5
= 1n
1 1 1 1
We can say that tn = n!
For example:
For FRANK’s name, tn = 5!
tn = 5 x 4 x 3 x 2 x 1
= 120
The formula for the number of combinations for a word where all the letters are different can be obtained, (as with the LUCY example on page 2) by simply writing out all of the permutations. As no letters are repeated, it is a straightforward formula.
There are a total of four letters to be placed into four different positions each time. When the first letter is placed, there are four different places to put it in. When the second letter is being placed, one position is already being taken up by the first letter, and so there are only three places where it can be positioned. When the third letter is being placed, there are only two places in which it can be placed, and the final letter must fill the final space.
In the formula, we start with the number four as there are four letters in the word, and at each stage of the arrangement process, there is one less available place in which to put the letters, so the multiplying number decreases at each stage
Therefore, where there are four letters to be placed, we can summarise it as:
tn = 4 x 3 x 2 x 1
or tn = 4!
(b). Investigating the number of different arrangements of letters in names with varying numbers of letters, where one letter is repeated once.
For example: words such as EMMA, or PETER
EMMA repeats the letter “M” once, and PETER repeats the letter “E” once, so that in each case one letter appears twice.
Investigation:
Name: ?
Number of letters: 1
Arrangements:
Note: There are no arrangements for this option as a name with
Only one letter cannot, by definition have a letter repeated.
Total Number of Arrangements: 0
Name: “JJ”
Number of letters: 2
Arrangements:
Total Number of Arrangements: 1
Name: “BOB”
Number of letters: 3
Arrangements:
Total Number of Arrangements: 3
Name: “EMMA”
Number of letters: 4
Arrangements:
Total Number of Arrangements: 12
Name: “PETER”
Number of letters: 5
Arrangements:
Total Number of Arrangements: 60
Hypothesis:
I think that in the case of names where 1 letter is repeated once (i.e. – the same letter appears twice), there is a formula of
Examples:
Name: “EMMA”
Number of letters: 4
Name: “PETER”
Number of letters: 5
As I explained earlier, when all of the letters of a word are different, the formula is
However, when a letter occurs twice, the total number of permutations is halved, and so the formula is obtained.
Investigating the number of different arrangements of letters in names with varying numbers of letters, where one letter is repeated twice.
For example: words such as BBBC, and CCCDE
BBBC repeats the letter “B” three times, and CCCDE repeats the letter “C” three times, so that in each case one letter appears three times.
Investigation:
Name: ?
Number of letters: 1-2
Arrangements:
Note: There are no arrangements for this option as a name with only
one or two letters cannot, by definition have a letter repeated twice.
Total Number of Arrangements: 0
Name: “AAA”
Number of letters: 3
Arrangements:
Total Number of Arrangements: 1
Name: “BBBC”
Number of letters: 4
Arrangements:
Total Number of Arrangements: 4
Name: “CCCDE”
Number of letters: 5
Arrangements:
Total Number of Arrangements: 20
Hypothesis:
Using a continuation of the formula I tried earlier, in cases where 1 letter is repeated twice (i.e. – the same letter appears three times), there should be a formula of
Examples:
Name: “BBBC”
Number of letters: 4
Name: “CCCDE”
Number of letters: 5
- This equation does not appear to work.
- The equation for the number of arrangements when one letter occurs twice did however work.
-
The equation was where n was the number of letters, and 2 the number of times a letter was repeated.
-
However, using the same idea was tried, but this did not work.
- It is possible that the denominator in these equations should also be factorial. 2! would still give the answer 2, making the equation appear correct when in fact it was not.
-
Therefore, for the names where 1 letter appears three times, I will try the equation
Examples:
Name: “BBBC”
Number of letters: 4
Name: “CCCDE”
Number of letters: 5
This equation does seem to work. The equation for the names where 1 letter occurs twice will be however this change would still give the same result.
Taking “CCCDE” as an example, we can see that the equation is
This is because there are a total of 5 different places the letters can be put into. For the first letter, there are 5 possibilities of places to position it. The second letter only has four places and so on. The equation is then divided by 3! Because, out of the five letters, three are the same. When one of these in used, only two are the same, and so on. Therefore, the equation is divided by (3 x 2 x 1), or 3!.
Example:
If we take a 5-letter word, with the letters A1, A2, A3, B, and C, it can be arranged as follows:
If however we remove the subscript numbers, certain combinations are repeated, when these are removed, we are left with the following arrangements:
This means that the resulting formula is
Investigating the number of different arrangements of letters in names with varying numbers of letters, where one letter is repeated three times.
For example: words such as BBBBC, or CCCCDE
BBBBC repeats the letter “B” three times, and CCCCDE repeats the letter “C” three times, so that in each case one letter appears four times.
Investigation:
Name: ?
Number of letters: 1-3
Arrangements:
Note: There are no arrangements for this option as a name with 1–3 letters cannot, by definition have a letter repeated three times.
Total Number of Arrangements: 0
Name: “AAAA”
Number of letters: 4
Arrangements:
Total Number of Arrangements: 1
Name: “BBBBC”
Number of letters: 4
Arrangements:
Total Number of Arrangements: 5
Name: “CCCCDE”
Number of letters: 6
Arrangements:
Total Number of Arrangements: 30
Hypothesis:
We would expect the above words all to follow the equation
As before, with 6 places to put the first letter in, 5 to put the second in, 4 to put the third in, and so on, you would get tn by multiplying 6 x 5 x 4 x 3 x 2 x 1. As there are four letters the same, the number of different possible arrangements gets smaller each time, and so the equation must be divided by 4!
Examples:
Name: “BBBBC”
Number of letters: 5
Name: “CCCCDE”
Number of letters: 6
Summary of Part 3b:
In names where 1 letter is repeated a certain number of times, the formula is:
where n = the number of letters in the word,
and R = the number of times the letter occurs.
Part 4:
Investigating the number of different arrangements of letters where more than one letter occurs more than once.
For example: words such as XXYY, XXYYXX, and XXXYYY
The above examples all have more than one letter that repeats more than one time.
Investigation:
Name: “XXYY”
Number of letters: 4
Arrangements:
Total Number of Arrangements: 6
Name: “XXXYYY”
Number of letters: 6
Arrangements:
Total Number of Arrangements: 20
Name: “XXYYY”
Number of letters: 5
Arrangements:
Total Number of Arrangements: 10
Summary of Part 4:
When there is only one letter repeated, the formula obtained was
Therefore, for words where more than one letter is repeated, I will try the formula
Where n = the total number of letters
And R1 = the number of times the first letter occurs
And R2 = the number of times the second letter occurs
Examples:
Name: “XXXYYY”
Number of letters: 6
This formula does not work. I think that the formula could perhaps be
Examples:
Name: “XXXYYY”
Number of letters: 6
Name: “XXYYY”
Number of letters: 5
This equation appears to be correct.
-
We have therefore arrived at the formula
-
The n! Is there as the total number of letters, as before. We previously decided that where one letter is repeated a certain number of times, the equation would be
- It is therefore logical to assume that where more than one letter is repeated, n! Should be divided by the total number of the first repeated letters multiplied by the total number of the second repeated letters.
