As you can see there are 120 different combinations for a 5-letter word. I can now see if there are going to be 120 if I use my formula.
5! = 1 x 2 x 3 x 4 x 5 = 120
My formula does work and it works for a 2,3,4 and five letter word so it is quite a reliable formula. This means that if I wanted to find out the number of different combinations for a 6 letter word then there will be: -
6! = 1 x 2 x 3 x 4 x 5 x 6 = 720
This means that there are 720 different combinations for a 6-letter word.
I can also work this out by using a different method. If I take “BLUCY” I can see that there are 5 different positions for each letter.
E.g. B L U C Y
If you take the five different positions and then multiply it by 24 (because this is the number of different combinations for each letter) you again get the answer of 120.
5 x 24 = 120 .
Now that I have found out a formula for “Lucy” and a word with all the letters different I need to find out a formula for “Emma”. It has come to my notice before I do anything at all that this name has 1 letter repeated and so is probably not going to be the same as “Lucy”. Again I will use the same method and start with a small word and work my way up.
Firstly I will take the name “Jj”: - jj
As you can see there is only 1 different combination for the name “Jj”
Now I will use the name “Bob”: - bob obb
bbo
As you can see the name “Bob” only has three different combinations.
Now I can use “Emma” and see how many different combinations there are for this name: - emma aemm mmae
eamm amme mmea emam amem meam mema mame maem
The name “Emma” has 12 different combinations. I can now put these results into another table and then see if I can recognise a pattern that will help me to find a formula.
A pattern can’t be recognised straight away so I can see if Factorial Notation will work again.
4! = 1 x 2 x 3 x 4 = 24
This as you can see does not work but what I have seen is that if i then divide the answer of 24 by 2 then i will get the answer that I am looking for.
24 / 2 = 12
I will now test with 2 and 3
2! / 2 3! / 2
= 1 x 2 / 2 = 1 x 2 x 3 / 2
= 2 / 2 = 6 / 2
= 1 = 3
This does work for all of these and will probably work for the rest. The formula is: -
“N! / 2”
I will now do this for the 5-letter word with two letters the same.
The word will be “bemma”. I will first use my formula to find it out and then I will test it by finding out all the different combinations.
N! / 2
= 5! / 2
= 1 x 2 x 3 x 4 x 5 / 2
= 120 / 2
= 60
I believe that when I work out all the combinations there will be 60 different combinations.
Let me see: - bemma ebmma mbema abemm bemam ebmam mbeam abmem beamm ebamm mbmea abmme bmema embma mbmae aebmm bmeam embam mbaem aembm bmmea emmba mbame aemmb bmmae emmab mebma ambem bmaem emabm mebam ambme
bmame emamb memba amebm
baemm eabmm memab amemb bamem eambm meabm ammbe bamme eammb meamb ammeb
mmbea mmbae mmeba mmeab mmabe mmaeb mabem mabme maebm maemb mambe mameb
As you can see from all of these different combinations there are in fact 60 which is what I predicted with my formula. So now I have evidence that my formula does work so it is most likely to work with and word with two letters repeated.
So if there was a 20-letter word with two of the letters the same then I could simply substitute into my formula and then find out the answer instead of trying to work them all out in your head.
N! / 2 = 20! / 2
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 x 15 x 16 x 17 x 18 x 19 x 20 / 2
= 12939908634132480000 / 2
= 6469954317066240000
I don’t think that anyone would enjoy working out this many different combinations. This is why I have found this formula, to make it easier to find out sums like these.
Now I will take it one level higher and use my own names. I will not use words with 2 letters the same but with 3 letters the same.
Firstly I will take “aaa”: - aaa
There is only 1 different combination for this.
Now I will take “aaab”: - aaab baaa aaba
abaa
There are four different combinations for this.
Now I will use a 5-letter word with 3 letters the same (aaabc): -
aaabc baaac caaab aabac baaca caaba aaacb bacaa cabaa
aabca bcaaa cbaaa aacab
aacba
abaac abaca abcaa acaab acaba acbaa
As you can see there are 20 different ways for a 5-letter word with three letters the same.
I can now put these results into a third table and again see if there is a pattern that will help me to create a formula.
I can’t see a formula that would work for this so I will try the other two that were used earlier.
Firstly there was the N! formula, So I will substitute in and then work it out to see if it is the correct formula for a word with 3 letters repeated.
5!
= 1 x 2 x 3 x 4 x 5
= 120
This formula does not work for a word with three letters repeated so I will now try the second formula, which was N! / 2
5! / 2
= 1 x 2 x 3 x 4 x 5 / 2
= 120 / 2
=60
This does not work either but it has helped me to notice a formula that may work. To get from the wrong answer of 60 to the right answer of 20 all you need to do is divide by 3. So really all you do is divide by 6 from the start. So what I am really doing is “N! / 6” but will this work for all of them. I will try it for a 4-letter word with three of the letters the same: -
N! / 6
= 4! / 6
= 1 x 2 x 3 x 4 / 6
= 24 / 6
= 4
This formula works for this as well as for 5. I think that this is a good formula and will work for all of them. So a 6-letter word with three letters the same will be: -
N! / 6
= 6! / 6
= 1 x 2 x 3 x 4 x 5 x 6 / 6
= 720 / 6
= 120.
Up till now I have found three different formulas that work for three different types of words either all different or with two or three repeated letters.
- The first formula was for a word of any amount of letters but all the letters are different. The formula for this is “N!”
- The second formula I found was for a word of an amount of letters but with two of one letter repeated. This formula was “N! / 2”
- The third and final formula that I have found was for a word with again any amount of letters but only one letter repeated three times. This formula was “N! / 6”
I may have found out all these formulas but it would make it much easier if I could find a formula that would work for all of them. This would mean that I would only have to know 1 formula instead of three different ones.
The formula that I have come up with for all three and for any word of any amount of letters with any amount of repeats is “N!/X!”.
An example of how this formula works is as follows: -
N!/X!
N! = Number of letters.
X! = Number of same letter.
E.g. If I was to find out all the different combinations for “aaabcd”
N!/X!
= 6! / 3 x 2 x 1
= 1 x 2 x 3 x 4 x 5 x 6 / 3 x 2 x 1
= 720 / 6
= 120
There are 120 different combinations for the word “aaabcd”
I will now explain how I found this formula: -
Lucy = 24 lucyz = 120 lucyzx = 720
Lucy lucyz lucyzx
24 120 720
4! 5! 6!
Lucc luccy luccyz
12 60 360
4!/2 5!/2 6!/2
I can see this works with 2 letters the same but what about 3.
Lccc lccc
24/3 = 8 cccl
4! / ? This is wrong clcc This is right
cclc We expect it to be “N!/3” but it isn’t.
The answer should be 4 and not 8.
We now need to put this into algebra.
We call: number of letters N
Number of letters the same X
Lucy = N/X we know that N is factorial so we now write it N!/X
Lucy = 4!/1 = 4! Or 1 x 2 x 3 x 4 or 24
Now I will try this for “Lucc”
Lucc = N!/x
= 4!/2
= 24/2
= 12
This works as well. Now lets try another one. We will try the one that did not work earlier “lccc” and we will see if we can work out a way to find the answer.
N!/x = 4!/? = 24/? =4
I have set this out as an algebraic equation and will see if I can work out the missing number.
What can divide 24 to give me the answer of 4?
Obviously it is 6.
At first I thought it was 3 but this gave me the answer of 8 which is wrong but when we use 6 it gives me the right answer.
So how did we get from 3 to 6?
What if we use factorial notation? E.g. 1 x 2 x 3 = 6
We thought you divide by the number of letters the same but we were wrong, it is in fact 6. Therefore if N is N! Why couldn’t X be X!. This would mean: - N!/X!
We can now test this on again the word “lccc”: -
N!/X!
= 4!/3!
= 24/6
= 4
Obviously this works on this equation, I will now try harder ones: -
Lcccc
N!/X!
= 5!/4!
= 120/24
= 5 Again this formula works. This formula in fact works for all the words that are used with any amount of letters and any amount of letters repeated.
Lets try it for all the Words that I used earlier when I made up singular formulas and see if this works for them.
Firstly we had “Lucy”: -
N!/X!
= 4!/1!
= 1 x 2 x 3 x 4 / 1
= 24 / 1
= 24
This formula works with this so now lets try it on “Emma”: -
N!/X!
= 4!/2!
= 1 x 2 x 3 x 4 / 1 x 2
= 24 / 2
= 12
Again we see that we get the same answer. So far so good, now lets try it for the third and final one “aaabc”: -
N!/X!
= 5!/3!
= 1 x 2 x 3 x 4 x 5 / 1 x 2 x 3
= 120 / 6
= 20
Again we get the right answer which is good news it works for all of these, so we now do not need the other three formulas that I made up earlier.
There is one alternative that I have discovered with this term.
If you have a word with two pairs of 2 letters the same then you will find that “N!/X!” does not work. But I have found a way round this. All you need to do is add a “Y!” this will make it work.
E.g. If we take the word “aabb” then we will try the formula that I have just discovered: -
N!/X!
= 4!/2! ?
I can see that there are two different letters repeated and I can’t substitute these into the equation. All I need to do is make the equation “N!/X!Y!
So I will try this and see if it works: -
N!/X!Y!
= 4!/2!*2!
= 1 x 2 x 3 x 4 / 1 x 2 * 1 x 2
= 24 / 2 * 2
= 24 / 4
= 6
This formula works with this as there are only 6 different combinations for this word. They are: - aabb baab abab baba
abba bbaa
as you can see there are only 6 different combinations.
If you are given a word such as “aabbcc” then you would simply add another letter to the term. It would become “N!/X!Y!Z!”.
I have tried to explain this in the simplest way possible and hope that it is understandable. My final formula is “N!/X!” and more letters can be added if there is more than one letter repeated.
