• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  • Level: GCSE
  • Subject: Maths
  • Word count: 1420

Dave's Dilemma

Extracts from this document...

Introduction

Maths Coursework 1. Investigate the number of arrangements of Dave's name......... DAVE DAEV DEAV DEVA DVAE DVEA ADEV ADVE AEDV AEVD AVED AVDE VDAE VDEA VEDA VEAD VAED VADE EVDA EVAD EDVA EDAV EAVD EADV There are 24 different arrangements for Dave's name. * For a two letter word (AT) , there are two arrangements. AT TA * For a three letter(CAT) word, there are six arrangements. CAT CTA ACT ATC TAC TCA * For a four letter word (LEAD), there are twenty-four arrangements. LEAD LEDA LAED LADE LDEA LDAE DEAL DELA DLEA DLAE DALE DAEL ADEL ADLE ALED ALDE AEDL AELD EALD EADL EDAL EDLA ELDA ELAD * For a five letter word (SOUTH), there are one hundred and twenty arrangements. SOUTH SOTUH SOHUT SOUHT SOHTU SOTHU SUOTH SUOHT SUHOT SUHTO SUTOH SUTHO STUHO STUOH STHUO STHOU STOHU STOUH SHTOU SHTUO SHUOT SHUTO SHOYU SHOUT OSUTH OSUHT OSHUT OSHTU OSTHU OSTUH OTSUH OTSHU OTHSU OTHUS OTUSH OTUHS OUTHS OUTSH OUSTH OUSHT OUHST OUHTS OHUTS OHUST OHSUT OHSTU OHTSU OHTUS UOSTH UOSHT UOHST UOHTS UOTSH UOTHS UTOHS UTOSH UTSOH UTSHO UTHOS UTHSO USTHO USTOH USOTH USOHT USHOT USHTO UHTSO UHTOS UHOTS UHOST UHSTO UHSOT THUOS THUSO THSUO THOSU THOUS TSHOU TSHUO TSUHO TSUOH TSOUH TSOHU TOSHU TOSUA TOUSH TOUHS TOHSU TOHUS TUHSO TUHOS TUSOH TUSHO TUOHS TUOSH HSOUT HSOTU HSTUO HSTOU HSUTO HSOUT HTSOU HTSUO HTUSO HTUOS HTOUS HTOSU HUOST HUOTS HUTOS HUTSO HUSTO HUSOT HOUST HOUTS HOTUS HOTSU HOSTU HOSUT The table below shows the number of arrangements for different length words. ...read more.

Middle

Number of letters Number of letters repeated Arrangements Calculation 2 2 1 2! divided by 2 3 2 3 3! divided by 2 4 2 12 4! divided by 2 5 2 60 5! divided by 2 6 2 360 6! divided by 2 7 2 2520 7! divided by 2 However, this formula was ineffective when trying to work out the number of arrangements for a word which has three letters the same in it, and all the rest different. AAAB AABA ABAA BAAA The word produced four arrangements, which is 24 divided 6(twenty-four being the number of arrangements of a four letter word with all the letters different, and six being the total number of arrangements of a three letter word and three is the number of letters repeated.) I worked out that the correct formula for this was....... n! 6 For example, if take the word ABACA, I found that there were twenty arrangements, these were...... ABACA ACABA AAABC AAACB BCAAA CBAAA AABCA AACBA ACBAA ABCAA ABAAC ACAAB BAAAC CAAAB BACAA CABAA CAABA BAACA AACAB AABAC This proves my formula. Below is a table to show this. Number of letters Number of letters repeated Arrangements Calculation 3 3 1 3! divided by 6 4 3 4 4! divided by 6 5 3 20 5! ...read more.

Conclusion

AAABBB BBBAAA ABBBAA AABBBA ABABAB BABABA BAAABB BBAAAB AABBAB ABBAAB BBAABA BAABBA ABAABB BABBAA AABABB BBABAA ABBABA BAABAB BABAAB ABABBA There are twenty arrangements of these letters. From these words I have discovered a formula that can be used to work out the number of arrangements for a word with more than one type of letter repeated, this formula is found from expanding on the formula used to find the number of arrangements for Emma's name (n!) . To find a word with more than one type of letter repeated you must multiply (y!) both 'y' values together. For example......... To find a the number of arrangements of an eight letter word with two letters and two lots of three letters the same (AAABBBCC) you would calculate 8!____ (3! x 3! x 2!) I then set about trying to find a formula for this. I quickly came to the decision that all that is needed is letters to represent the number of repeated letters. This gave me the formula: ____n!____ (x! X y! X z!) ('x' represents the number of repeated A's, 'y' represents the number of repeated B's and 'z' represents the number of repeated C's) To test my formula I used a five letter word (AABBC), with two lots of letters the same. This gives me the following number of arrangements.......... CABBA CBAAB CAABB CBBAA CBABA CABAB AABBC AACBB AABCB ABABC ABBCA ABCBA ACBBA ABBAC ACABB ACBAB ABCAB ABACB BBAAC BABAC BBCAA BAACB BACAB BCAAB BAABC BCBAA BCABA BACBA BBACA BABCA ...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. Emma's Dilemma

    ! = 5 X 4 X 3 X 2 X 1 = 120 30 2 ! X 2 ! X 1! 2 X 2 X 1 4 For this equation, there seems to be 30 different combinations of the name SASHA.

  2. GCSE maths coursework: Emma's dilemma

    Let's try 7 fig, with 3 same number, and 4 same number. I expect the total arrangement is a=(1*2*3*4*5*6*7)/(3*2*1*4*3*2*1)=35 1112222 1222211 2222111 2211212 1121222 1222121 2221211 2211221 1122122 1222112 2221121 2212112-----10 arrangements 1122212 1221221-----15 arrangements 2221112 2212121 1122221 1221212 2211122 2212211 1212221 1221122 1212212 1212122 1211222 2111222 2112221 2121221 2122211

  1. I have been given a problem entitled 'Emma's Dilemma' and I was given the ...

    There is also a letter (A) which comes up 2 times in the word. Therefore I get 2! which is equal to 2. So I get ( 1 ) x ( 1 ) x ( 1 x 2 ) = 1 x 1 x 2 = 2 So we get:

  2. Emma's Dilemma

    is repeated. If we look at XXYY for example, there are 6 different arrangements because both Xs are the same and both Ys are the same. XXYY XYXY XYYX YXYX YYXX YXXY But if the Xs and Ys were different there would be more arrangements.

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

    letters, with one repeated twice ), but to make sure that it didn't matter which letters were used, I decided to try them again. I found out that my analysis was correct, and that the changing of letters didn't affect the number of different arrangements for four letters, with one repeated twice.

  2. Emma’s Dilemma.

    Seeing this, it also became obvious of the links between each name. To find the number of permutations for a name, you take the number of letters in the name and multiply that by the number of permutations for the previous name.

  1. To investigate the various arrangements that could be made with different names or words. ...

    TRAI 18) TIRA 19) ARIT 20) ATIR 21) AITR 22) ARTI 23) ATRI 24) AIRT I have created 24 different arrangements using the name RITA. This is the same amount of arrangements found when rearranging LUCY's name. This is because they are quite similar, as they both have four letters and each letter is different from each other.

  2. GCSE Maths Project – “Emma’s Dilemma”

    In the formula, we start with the number four as there are four letters in the word, and at each stage of the arrangement process, there is one less available place in which to put the letters, so the multiplying number decreases at each stage Therefore, where there are four

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