# Emma&#146;s Dilemma.

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Introduction

Emma's Dilemma 1) 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). ...read more.

Middle

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 1,1 2,1 2,2 3,2 3,3 4,3 No. of letters! 2 6 24 120 720 5040 Combinations 2 3 6 10 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. ...read more.

Conclusion

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). ...read more.

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