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  • 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.

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