Mathan Navaratnam 11F 24.10.2001
Mathematics Coursework
Emma's Dilemma
The different arrangements for the name Emma are as follows:
Emma emam eamm mmae
Mmea meam mema mame
Maem amme amem aemm
There were a total of twelve combinations of the rearrangement of the name Emma. The name Emma consists of four letters where three of them being different, the two like letters being m.
Emma's friend however, Lucy, name consists again of four letters but this time all the letters are different. The number of arrangements for the name Lucy is as follows:
Lucy ucyl cylu ycul
Luyc ucly cyul yclu
Lcuy ulcy culy yulc
Lcyu ulyc cuyl yucl
Lyuc uycl clyu ylcu
Lycu uylc cluy yluc
This time there were twenty-four different possibilities in rearranging the name Lucy. This is significant, as it is exactly double the number of arrangements of the name Emma where there were 4 letters but where only 3 were different.
Looking back at the table above I think it would be useful to note how to predict the number of arrangements. So far I have worked out for names with four letters. If I am to use the name Lucy then I notice in my above table that there are six combinations for the rearrangements if the rearrangement was to say the letter l, six combinations with rearrangements starting with the letter u and so forth.
Thus 4 (number of different letters) multiplied by 6 (number of combinations per starting letter) = 24 (total combinations).
Now that I have investigated the number of possible rearrangements involving four letters with no similar letters and one similar letter, I should investigate the number of possibilities with tow and three similar letters. Obviously four similar letters will only have one possible combination.
The next step is to investigate the number of possibilities there will be in rearranging a five-letter word such as Clive. From the previous experiment with four different letters there were 24 combinations. Therefore with a five-letter word there should be 120 combinations, as there would be 24 combinations with 4 of the letters and the fifth letter would just be placed in front. As there are five different letters which could be placed at the front then the total number of combinations would be 5 (the number of letters) X 24 (the number of possibilities of rearrangements ...
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The next step is to investigate the number of possibilities there will be in rearranging a five-letter word such as Clive. From the previous experiment with four different letters there were 24 combinations. Therefore with a five-letter word there should be 120 combinations, as there would be 24 combinations with 4 of the letters and the fifth letter would just be placed in front. As there are five different letters which could be placed at the front then the total number of combinations would be 5 (the number of letters) X 24 (the number of possibilities of rearrangements for the rest of the letters) = 120 combinations.
For example I will show all the possible combinations there will be with the starting letter of C
Clive cliev clvie clvei
Cleiv clevi cvlie cvlei
Cveil cveli cviel cvile
Ceilv ceivl celvi celiv
Cevil cevli cielv cievl
Cilve cilev civle civel
The above 24 combinations are all the possible combinations with the starting letter of c. As there are a further 4 possible starting letters then there are still another 96 combinations (4 X 24) plus the above 24 combinations which gives the total of 120 combinations for a five letter word where all the letters are different, this is what I predicted earlier.
It is clearly evident that as the number of different letters increase, the total number of combinations will also increase dramatically and it would not be practical to write out the all the combinations. So I will try and deduce a formula from my previous findings. If I look at the combinations with all different letters then with four different letters, Lucy, there were 24 combinations, for five different letters, clive, there were 120 different combinations. It is obvious that four one letter, such as a, there will be one combination and for two different letters, such as jo, there will be two different combinations (jo, and oj) and for a word with all three different letters, such as joe there will be six different combinations (joe, jeo, eoj, ejo, oej, oje). So for one different letter there is one combination, with two letters there is 2 combinations, for three letters there is 6 combinations, for four letters there is 24 combinations and for five letters there is 120 combinations. This is interesting as for one letter there is one combination (1), for two letters there is two combinations (1 X 2), for three combinations there were six combinations, (1 X 2 X 3), for four letters there were 24 combinations (1 X 3 X 4) and for five letters there were 120 combinations (1 X 2 X 3 X 4 X 5).
That means by following this pattern, for a six letter word where all the letters are different there should be 720 combinations, (1 X 2 X 3 X 4 X 5 X 6). This would make sense as there would be six starting letters and the remaining five different letters can be rearranged 120 times, as we discovered earlier. So for each starting letter there will be 120 combinations and there are 6 different starting letters so there will be 120 X 6 = 720 combinations. 720 = (1 X 2 X 3 X 4 X 5 X 6) which follows the formula which I found earlier. (1 X 2 X 3 X 4 X 5 (the formula for five different letters) X 6 (number of different stating letters). This is expressed as factorial or !. Factorial represents the calculation of 1x2x3x4x5x6x7... etc. For example 6! = 720, 5! = 120 and so on. So for words where all the letters are different I can now calculate the number of different combinations.
Total letters in word (all different)
Number of possible combinations
2
2
3
6
4
24
5
20
6
720
7
5040
55
.269640335 x 1073
n
n!
Now that I have explained the patter of words where every letter is different I need to investigate into words where some of the letters are similar such as Emma. If all the letters in Emma were different then there would be 24 different combination but there are two similar letters. The total number of possible combinations for Emma is 12 and not 24 which means the factorial rule which was used for all different letters earlier either does not apply or applies as well as accompanied by something else.
For ease of understanding I will replace letters simply with x's and y's. For example I will investigate into a four-letter word where there are only two different letters. This word could be noon where the n's and o's are replaced by x's and y's.
So the possible combinations for xxyy are as follows:
Xxyy xyxy yxxy
Xyyx yxyx yyxx
There are 6 possible arrangements of xxyy.
Now I will investigate into a five-letter word with a total of two different letters. This will be represented as xxxyy:
Xxxyy xxyxy xyxyx xyxxy xxyyx
Xyyxx yyxxx yxxxy yxyxx yxxyx
There are 10 different possible rearrangements for xxxyy.
Finally I will investigate into a five-letter word where there is four similar letters and one different. This will be represented as xxxxy:
Xxxxy xxxyx xxyxx xyxxx yxxxx.
There are a total of five possible rearrangements for xxxxy.
In the case of xxxyy, in a total of a five-letter word there were 3 x's and 2 y's. Also out of the total of ten combinations for xxxyy, five of those combinations started with x and the other five started with y. As the rest of this investigation involves the factorial calculation it only seems logical that I investigate further into the factorial calculations. If the letters were all different then a five-letter word would have 120 combinations but in the case of xxxyy where there were similar letters there were 10 combinations. There were 3 x's and 2 y's. There should be a link between the number of x's and y's present involving the factorial calculation with the total number of letters in the word which tells us the number of combinations for the word, say xxxyy. There were 3 x's and 3! = 6. There were 2 y's and 2! = 2. There were a total of five letters in the word a 5! = 120. From this we can deduce that 3!x2! = 12. This is significant as 120 (which is 5!) divided by 12 gives us 10 which is exactly the number of combinations there was for xxxyy. Therefore 5! / 3!2! = 12. This clearly works in the case of xxxyy.
We can replace numbers in this formula to letters. Therefore the formula would be n!/x!y!
To test whether this is the correct formula I will apply it to the above examples.
So for xxxyy; there are 3 x's present and 2 y's present. Therefore 1x2x3x4x5 / (1x2x3) x (1x2) = 120/12 = 10.
For xxxxy, there are 4 x's present and 1 y present. Therefore 1x2x3x4x5 / (1x2x3x4) x (1) = 120 / 24 = 5.
For xxyy there are 2 x's present and 2 y's present. Therefore 1x2x3x4 / (1x2) x (1x2) = 24 / 4 = 6
This formula should also work where every letter of the word is different. So for a word such as vwxyz there is only one of each letter present. Therefore 1x2x3x4x5 / (1)x(1)x(1)x(1)x(1) = 120 / 1 = 120.
The above calculations prove that my formula is correct.
