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Emma's Dilemma

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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): 1- ? (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 ?. ...read more.


(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. 1- ? (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: 1- ? (aaa) = aaa = 1. 2- ? (aaas) = aaas, asaa, aasa, saaa = 4. ? for 3 and 4 letter words with no repeated letters: 1- ? (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. ...read more.


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!. 120 / 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: 1- 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 ...read more.

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