MOHAMED KHALIFA MATHEMATICS
INTRODUCTION:
For mathematics, we were given the Coursework of investigating "Emma's Dilemma." This Coursework is set up in a number of steps which builds up to a formula in which one could enter any number of letters, with or without repeated letters, and come out with the number of possible permutations in which these words could be expressed. For this reason, I decided that I will not follow the steps and begin by investigating what is required, which is come up with the formula. I shall begin my task by looking at the possible number of ways in which I could write a word which has no letters repeated in it. I shall express the phrase "possible number of words" with the symbol ?.
WORDS WITH NO REPEATED LETTERS:
I shall start by looking at such words starting from single lettered words (Most of my words will not make sense):
- ? (a) = 1
2- ? (ed) = ed, de = 2
3- ? (cat) = cat, cta, tac, tca, act, atc = 6
4- ? (lucy) = (lucy, luyc, lcuy, lcyu, lycu, lyuc) * 4 = 24.
Notice how I multiplied the last example by 4 to get the ?. This is because I have noticed something as I was working out the ?. I have found that I was going to repeat everything I was doing 4 times. Similarly, I have realised that I was doing every thing 3 times for the second letter, twice for my third letter and once for my last. Hence, I could say that to work out the ? for a four letter word, I am merely multiplying 4*3*2*1 and this gives me 24, which is the correct answer. I could express this as 4!. Could I however say this for words made up of a different amount of letters? I shall test this.
I have already worked out the ? of 1, 2 & 3 letter words so I will thus test what I have found on them.
Will the ? of 3, 2, 1 be 3!, 2!, 1! ?
(i) 3! = 3*2*1 = 6. The expression is successful for 3.
(ii) 2! = 2*1 = 2. The expression is successful for 2.
(iii) 1! = 1. The expression is successful for 1.
Having tested the expression, I could therefore say that it is a success. I could also generalise this expression and say that if X is the number of letters in any word, then ?(X) = X!.
WORDS WITH REPEATED LETTERS:
Having found that ?(X) = X! where X is the number of letters in a word with no repititions, ...
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(i) 3! = 3*2*1 = 6. The expression is successful for 3.
(ii) 2! = 2*1 = 2. The expression is successful for 2.
(iii) 1! = 1. The expression is successful for 1.
Having tested the expression, I could therefore say that it is a success. I could also generalise this expression and say that if X is the number of letters in any word, then ?(X) = X!.
WORDS WITH REPEATED LETTERS:
Having found that ?(X) = X! where X is the number of letters in a word with no repititions, I could say that this expression could have a relationship with ? in words with letters repeated. To be more specific, I mean that a 4 letter word with no repeated letters, could be written in more combinations than that of which has repeated letters e.g. Lucy could be written in more ways than Emma. This is because the letter "M" in Emma could be written in only one way so although there are 2 M's in Emma, once one is used in one form, the other could not be used again in the same form as it will count as repitition. Is it therefore possible to say that ? in repetitive words will equal to X! divided by abc. This does not mean that the divisible (abc) will be constant. I question this because I know that there will be a relationship between ? in non repetitive and repetitive words and since it will be less, I thing that a division will take place. I will investigate this.
I will start by looking for a relationship between the ? of words with 2 of the same letters repeated and the ? of words made up of the same number of letters with no letters repeated. I will do this by testing my prediction. I will test my prediction by keeping the number of repeated letters in the word the same but varying the length of the word.
? for 2, 3, and 4 letter words with 2 repeated letters:
- ? (aa) = 1.
2- ? (saa) = saa, asa, aas = 3.
3- ? (saab) = saab, sbaa, saba, baas, bsaa, basa, aabs, asab, absa, aasb, abas, asba = 12
? for 2, 3 and 4 letter words with no repeated letters:
- ? (ed) = 2
2- ? (cat) = 6 (SEE "WORDS WITH NO REPEAED LETTER SECTION"
3- ? (lucy) = 24 FOR WORKING OUT.)
My prediction states that:
X! / abc = ? of words with repeated letters. I will test this to find abc by rearranging the formula. abc = X! / ? of words with repeated letters.
- ? (ed) / ? (aa) = abc ? 2 / 1 = abc, abc = 2.
2- ? (cat) / ? (saa) = abc ?6 / 3 = abc, abc = 2.
3- ? (lucy) / ? (saab) = abc ? 24 / 12 = abc, abc = 2.
I find that I am getting 2 for abc. 2 is therefore constant. Does this 2 stay constant if the same letter is repeated 3 times in the word. I shall investigate this by testing my prediction on the ? of words with the same letters repeated thrice but the length of the word varies.
? for 3 and 4 letter words with 3 repeated letters:
- ? (aaa) = aaa = 1.
2- ? (aaas) = aaas, asaa, aasa, saaa = 4.
? for 3 and 4 letter words with no repeated letters:
- ? (cat) = 6 (SEE "WORDS WITH NO REPEAED LETTER SECTION"
2- ? (lucy) = 24 FOR WORKING OUT.)
My prediction states that:
X! / abc = ? of words with repeated letters. I will test this to find abc by rearranging the formula. abc = X! / ? of words with repeated letters.
- ? (cat) / ? (aaa) = abc ? 6 / 1 = abc, abc = 6.
2- ? (lucy) / ? (aaas) = abc ? 24 / 4 = abc, abc = 6.
I find that I am getting 6 for abc. 6 is therefore constant. If I recall everything I have achieved with repeated letters I find that:
- ? of words with two of the same letters repeated in them = 2.
2- ? of words with three of the same letters repeated in them = 6.
In writting these two statements, I have noticed that that 2 = 2! and 6 = 3!. I will therefore predict that abc for words where four of the same letter is repeated is equal to 4!. I will test this.
? of 4 letter word with 4 of the same letters repeated.
- ? (aaaa) = aaaa = 1.
? of 4 letter word with no repeated letters.
- ? (lucy) = 24
My prediction states that:
X! / abc = ? of words with repeated letters. I will test this to find abc by rearranging the formula. abc = X! / ? of words with repeated letters.
? (lucy) / ? (aaaa) = abc ? 24 / 1 = abc, abc = 24.
From what I have just achieved, I think that I could deffinately state that ? of words with one letter repeated more than once, is equal to the factorial (!) of the number of letters in the word divided by the factorial of the number of times a letter has been repeated in the word. This can be expressed in a formula:
? of words with one letter repeated more than once = X! / A!.
X= no. of letters in the word.
A= no. of times one letter has been repeated.
What is the situation however when we one to find out the ? of a word where more than one of its letters has been repeated? The solution is expressed in this formula:
? of words with more than one letter repeated more than once = X! / A! * B!.
X= no. of letters in the word.
A= no. of times one letter has been repeated.
B= no. of times another letter has been repeated.
In order for me to work out the last formula, I did not have to carry out investigations, as it makes sense. B! was divided by X! because it was in the same situation as A!. It was multiplied by A!, as the ? was going to be less with more than one repeated letter and by multiplying, we make the bottom heavier thus the final calculation will be less. I will prove it however:
? of a 5 letter word with more than one repeated letter such as aasss = aasss, asass, assas, asssa, sssaa, ssasa, ssaas, sassa, saass, ssaas = 10.
By using the formula above, we will find that ? of a 5 letter word with more than one repeated letter such as aasss will equal to 10.
X! = 120 ( as X is a 5 letter word)
A! = 2 ( as A is repeated twice)
B! = 6 ( as S is repeated thrice)
I shall substitute the letters in the formula I have above with the numbers I found for X!, A!, B!.
20 / 2*6 = 10. The formula is successful
This formula is succesful, then it should work for words with 2, 3, 4, 5 etc. For example, the ? of a word with three different repeated letters (aabbcc) is:
? (aabbcc) = 6! / 2!*2!*2! = 120
Simillarly, the ? of a 12 letter word, with 4 repeated letters (aaabbbbcccdd) is:
? (aaabbbbcccdd) = 12! / 3!*4!*3!*2! = 277200
CONCLUSION:
In this coursewok, I have found the formulas which allow us to find the ? of words with or without repeated letters. The formulas are:
- Formula to find the ? of words where there are no letters repeated: X!.
2- Formula to find the ? of words where there is one letter repeated more than once: X! / A!
3- Formula to find the ? of words where there is more than one letter repeated more than once: X! / A! * B!. In this formula, I have shown the ? of the words, considering that only 2 letters, A & B, have been repeated more than once. This formula will work for any number of letters repeated, by multiplying the ! of the frequency of the repeated letters to the bottom of the fraction in the formula.
MATHEMATICS COURSEWORK
PERMUTATIONS
BY:MOHAMED KHALIFA
