• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  • Level: GCSE
  • Subject: Maths
  • Word count: 1575

Emma’s Dilemma.

Extracts from this document...


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.


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.


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.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Emma's Dilemma essays

  1. Arrangements for names.

    for each letter. As that seemed to work, I had a go at trying to work out a logical universal formula. I came up with; The total number of letters factorial, divided by the number of a's, b's ect factorised and multiplied Formula for emma= 4!/1!x1!x2!=48 /4=12 For example: A

  2. Emma's Dilemma

    of letters - 1 ) X ( No. of letters - 2 ) X etc... combinations ( times a letter X ( times a different letter has been repeated ) has been repeated ) From this rule, all that is needed to be done, is to introduce a small mathematical function, called "FACTORIALS ( !

  1. To investigate the combination of arrangement of letters in Jeans name and then for ...

    The number of combinations is 10, which would mean that you needed to divide 20 by 2 again. So it would be the calculation: 120 / 3 / 2 / 2. I though that you had to divide by the number of letters there are the same for each one, for example.

  2. Emma's Dilemma Question One: Investigate the number of different arrangements of the letters

    XYXYYY XYYXYY XYYYXY XYYYYX YXXYYY Total: YXYXYY 15 YXYYXY YXYYYX YYXXYY YYXYXY YYXYYX YYYXXY YYYXYX YYYYXX Results for Two X's and No. of Y's: Number of Y's Number of different combinations: 0 1 1 3 2 6 3 10 4 15 Rule: To find the number of all the different

  1. permutations & combinations

    In this word we are taking all the letters separately so we get 120 variations. In the other word, there are 5 letters but 3 similar for one letter and two similar for the other. So the formula for A1A2A3B1B2 will be 5!

  2. I am doing an investigation into words and their number of combinations. I will ...

    fits the graph. Therefore the formula for words with 3 letters the same is: n!/3 Prediction At this point it is becoming clearer to me the pattern between all the different numbers and formulae. It is a basic statement that all the formulae are involved in n!

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work