Emma's Dilemma
) I will methodically write down all the combinations of the letters of Emma's name:-
MEMA EMMA AMME
MEAM EMAM AEMM
MAEM EAMM AEMM
MAME
MMAE
MMEA
As you can see from the above the results are: 4 letters, 2 the same, 12 combinations. (Remember there are two letters that are the same and therefore this affects the amount of combinations, as we will see later).
2) With long names it is hard to work out the number of different combinations and ideally we need to find a formula instead of having to right them all out.
Firstly, it would probably be easier if we started with a simple name with no repeating letters, not to complicate things, and then I will simply work up:
Name: AD Two letters
AD DA
Combinations: 2
Name: JON Three letters
JON OJN NJO
JNO ONJ NOJ
Combinations: 6
Name: LUCY Four letters
LUCY UCLY YLCU CULY
LUYC UCYL YLCU CUYL
LYUC ULCY YUCL CLYU
LYCU ULYC YULC CLUY
LCUY UYLC YCLU CYUL
LCYU UYCL YCUL CYLU
Combinations: 24
Name: SIMON Five letters
SIMON SOIMN SMION SNIOM
SIMNO SOINM SMINO SNIMO
SINMO SONMI SMNOI SNMIO
SIOMN SONIM SMNIO SNMOI
SINOM SOMIN SMOIN SNOMI
SIONM SOMNI SMONI SNOIM
In this case, the name I chose was too long but with 24 combinations with 'S' first, there are 120 combinations overall (24x5).
Combinations: 120
From the above I can work out a table of results to enable me to work link the number of letters all different with its amount of combinations.
No. of letters, all different
2
3
4
5
6
7
8
9
0
Combination
2
6
24
20
?
?
?
?
?
From the above table I can recognise that the number of combinations ina word where all the letters are different is the total number of letters factorial. (!)
No. of combinations = No. of letters !
C = L!
Where C= number of combinations and L = number of letters.
The factorial function is the formula that multiplies the number with all of its previous numbers down to 1 i.e. 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.
Therefore using the information the table above would look like this:-
No. of letters, all different
2
3
4
5
6
7
8
...
This is a preview of the whole essay
No. of combinations = No. of letters !
C = L!
Where C= number of combinations and L = number of letters.
The factorial function is the formula that multiplies the number with all of its previous numbers down to 1 i.e. 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.
Therefore using the information the table above would look like this:-
No. of letters, all different
2
3
4
5
6
7
8
9
0
Combination
2
6
24
20
720
5040
40320
362880
3628800
3/4) As noted upon before, the name EMMA has two letters the same and so it has fewer combinations for example:-
XYY XYZ
YXY XZY
YYX ZXY
ZYX
YZX
YXZ
3 combinations 6 combinations
(should be 6 comb.)
We found a formula before C = L!, but it doesn't work for names where there are two letters the same or even three and so therefore I need to work out another formula. As from before I expect that the formula will involve the factorial function and I will start as I did before with simple formations, although this time I will use letters and not names.
XY XYY XYYY XYYYY XYYYYY XYYYYYY
YX YXY YXYY YXYYY YXYYYY YXYYYYY
YYX YYXY YYXYY YYXYYY YYXYYYY
YYYX YYYXY YYYXYY YYYXYYY
YYYYX YYYYXY YYYYXYY
YYYYYX YYYYYXY
YYYYYYX
2 comb. 3 comb. 4 comb. 5 comb. 6 comb. 7 comb.
From the table below it is clear that the total number of combinations is just the same as the total number of letters. This is right but this is only right when there is only one repeating letter. Therefore I need to investigate the number of combinations when you have a different number of repeating letters.
No. of letters
2
3
4
5
6
7
No. of same
2
3
4
5
6
Combinations
2 (2)
3 (6)
4 (24)
5 (120)
6 (720)
7 (5040)
It is clear that the formula I am trying to find out isn't the total number of letters factorial (shown in brackets above). Therefore, the formula has to be total number of letters minus something, add something, multiply something or divide by something. Now I will investigate the following possibilities, which will eventually enable me to work out a suitable formula to find out the number of different combinations when there are different amounts of repeating letters e.g. (XXXYY)
XY
YX 2 combinations
XXY
XYX
YXX 3 combinations
XXYY
XYYX
XYXY
YXYX
YYXX
YXXY 6 combinations
XXXYY
XXYXY
XYXXY
YXXXY
XXYYX
XYXYX
XYYXX
YXXYX
YXYXX
YYXXX 1O combinations
XXXYYY XXYYYX
XXYXYY YXXYXY
XYXXYY YXYXXY
YXXXYY YXXYYX
XYXYXY YXYXYX
XYYXYX YYYXXX
XYYXXY YYXXXY
XYYYXX YYXXYX
XYXYYX YXYYXX
XXYYXY YYXYXX 20 combinations
XXXXYYY XXYYXYX YYXYXXX XXYYXXY
XXYXYYX XXYYYXX YYXXYXX XYXXXYY
XYXXYYX XXXYYYX YYXXXYX XYXXYXY
YXXXYYX YXXXYYX YYXXXXY XYXXYYX
XYXYXYX YXXYXYX XXYYXXY XYXYXXY
XYYXYXX YXXYYXX XXXYXYY XYYXXXY
XYYXXYX YXYXYXX XXXYYXY YXYXXYX
XYYYXXX YXYYXXX XXYXXYY YXYXXXY
XYXYYXX YYYXXXX XXYXYXY 35 combinations
Letters
2
3
4
5
6
7
No. of same
,1
2,1
2,2
3,2
3,3
4,3
No. of letters!
2
6
24
20
720
5040
Combinations
2
3
6
0
20
35
As you can see the row "No. of letters!" has to equal the row, "Combinations."
In order to do this I have to think of a way that might connect the two sets of data, i.e. +,-,x,/. Therefore, I will try to work out the formula using a trial and error method.
Currently, I know the No. of combinations and all I need to do is to find a formula that satisfies them. I will use the example, XXXYY. In the case of XXXYY it has 10 combinations.
Possible formulas
. Comb. = No. of letters! - (No. of X + No. of Y)
Comb. = 120 - (3+2)
= 115
This formula doesn't work.
2. Comb. = No. of letters! - (No. of X x No. of Y)
Comb. = 120 - (3x2)
= 114
This formula doesn't work.
3. Comb. = No. of letters!/(No. of X + No. of Y)
Comb. = 120 / (3+2)
= 24
This formula is also wrong, although it is getting closer to the answer. Could this have anything to do with the division sign coming into play?
4. Comb. = No. of letters! / (No. of X x No. of Y)
Comb. = 120 / (3x2)
= 20
Once again this doesn't work but we are getting closer!
5. Comb. = No. of letters! - (No. of X! + No. of Y!)
Comb. = 120 - (6+2)
= 15
With the introduction of the factorial function in the second half of the equation the formula is slowly getting nearer to the correct answer.
6. Comb. = No. of letters! / (No. of X! + No. of Y!)
Comb. = 120 / (6+2)
= 15
This doesn't work.
7. Comb. = No. of letters! / (No. of X! / No. of Y!)
Comb. = 120 / (6/2)
= 40
This also doesn't work!!
8. Comb. = No. of letters! / (No. of X! x No. of Y!)
Comb. = 120 / (6x2)
= 10
At last, this formula seems to work! To make sure I am going to use it on another set of results, XXXYYY, which has 20 combinations.
Comb. = 720 / (6x6)
= 20
This formula has worked again so I will try it with another example, XXXXYYY, which has 35 combinations.
Comb. = 5040 / (24x6)
= 35
Therefore the number of combinations in a name where there are two repeating letters can be worked out by using the factorial function on the number of letters, and then by dividing that result by the number of X's! multiplied by the number of Y's! :-
Combination = Letters! / (No. X's! x No. Y's!)
C = L! / (X! x Y!)
Having found the formula, which is stated above, I am now going to try some combinations that have three repeating letters to prove that the formula still works.
Presumably the formula should be the following, for combinations which have three repeating letters:-
Comb. = letters! / (No. of X! x No. of Y! x No. of Z!)
XYZ has 6 combinations and so:-
Comb. = 6 / (1x1x1)
= 6
This seems to work for this basic combination but if we make it more difficult does the formula still work?
XXYZ ZYXX
XXZY XYZX
XZXY XYXZ
XZYX YXZX
ZXYX YXXZ
ZXXY YZXX 12 combinations
Comb. = 24 / (2x1x1)
= 12
Once again this formula seems to work, though finally I will use XXYYZ
XXYYZ ZXYXY XYXYZ YXYXZ XYYZX YXYZX
XXYZY ZYYXX XYXZY YXXYZ XYZXY XZYYX
XXZYY ZXYYX XZYYX YZXXY XYZYX YYXZX
XZXYY ZYXYX XZYXY YZXYX YYZXX YXZXY
ZXXYY XYYXZ YYXXZ YZYXX XXZYY YXZYX
The letters XXYYZ have 30 different combinations. Now I have to try this with the formula, C = L! / (No. X! x No. Y! x No. Z!)
Combination = 5! / (2! x 2! x 1!)
Combination = 120 / 4
Combination = 30
This appears to agree with the formula's answer, which means if there were a pattern of one thousand different symbols repeating themselves then the formula would simply be:-
Comb. = Letters! / (No of A's! x No. of B's! x No. of C's! x No. of D's!......etc)
This formula is known as a universal formula and has been derived from my original one.
In conclusion, I have worked out a formula for working out the different amounts of combinations in a name that has no repeating letters, (C = L!). As well as this, I have worked out the formula for an arrangement of letters for two or more repeating letters, (C = L! / (No. A! x No. B! x No. C! ...etc).
