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GCSE: Emma's Dilemma

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1. Emmas dilemma

All the names I investigate will all have different letters. Name: Jo Total number of letters: 2 jo oj Jo has 2 different combinations. Name: Ian Total number of letters: 3 ian ina ani ain nai ain Ian has 6 different combinations. I have realised that the total number of combinations is the number of letters multiplied by the previous number of combinations. Name: Lucy Total number of letters: 4 Previous number of combinations: 6 Prediction: 4 x 6 = 24 lucy luyc lcyu lcuy lycu lyuc ulcy ulyc uylc uycl ucly ucyl cluy clyu culy cuyl cyul cylu yluc ylcu yclu ycul yucl yulc As i predicted 24 combinations.

• Word count: 755
2. Maths GCSE Coursework: Emma's Dilemma

After two letters were written, only two can be left. These letters came next. The next letter was the last. Now I have finished finding the combinations of L. I then used the same method to find the combination of the other ''starters'' (U, C and Y). Another way of displaying this method is shown below: By using this method, I know that I have discovered all combinations of the word Lucy. There were 24 combinations in total. Combination of Emma I used the same method for finding the combination of Lucy as I did to find Emma.

• Word count: 2245
3. To explore and find a relationship between the number of letters in a word and the number of arrangements of the letters there are.

>Ktï¿½"E1/4ï¿½ï¿½ &ï¿½\$:ï¿½ï¿½ï¿½ï¿½Eï¿½#:1/2ï¿½rdczï¿½ï¿½yï¿½1/2 +G6ï¿½ï¿½ï¿½ï¿½ï¿½i ï¿½v+a×²(r)ï¿½ï¿½ï¿½ï¿½"ï¿½ï¿½aYvMï¿½ï¿½ (r)ï¿½S@eï¿½ï¿½-%ï¿½Gï¿½?e3ï¿½ï¿½ï¿½ï¿½=Õ°GA}Jï¿½ï¿½Qï¿½ï¿½ï¿½ï¿½/ï¿½Oï¿½: ï¿½~ï¿½ï¿½3/4Yï¿½ï¿½l>ï¿½51 ï¿½7ÞSï¿½ï¿½yÊBdï¿½N-ï¿½ï¿½ï¿½1/4ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½Qï¿½Cï¿½ï¿½ï¿½"ï¿½L"ï¿½6zï¿½ï¿½ï¿½']&aï¿½F/7aï¿½ï¿½Øv ï¿½-jHï¿½+"(r)ï¿½ ï¿½42_ï¿½5Xï¿½ï¿½ï¿½ï¡­"ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½"ï¿½ ï¿½ï¿½ Nï¿½B"ï¿½Eï¿½ï¿½T&]ï¿½ ï¿½2ï¿½7 gï¿½ï¿½*ï¿½ï¿½bï¿½ï¿½0(tm)oï¿½^Z3/4ï¿½ï¿½hï¿½Õï¿½ï¿½ï¿½`ï¿½8="ï¿½Fï¿½7ï¿½|ï¿½aï¿½L9"ï¿½2e3ï¿½8lï¿½ï¿½7ï¿½N-ï¿½Øï¿½"ï¿½dj%S.{ï¿½Nï¿½xï¿½Lï¿½S.{ï¿½Nï¿½ Íºï¿½|ï¿½gï¿½"ï¿½2ï¿½y(ï¿½3/4ï¿½')ï¿½g=Iï¿½Z;%cï¿½4 ï¿½ï¿½)-gï¿½ï¿½ï¿½=2ï¿½i2ï¿½8ï¿½ï¿½7cï¿½Y7ï¿½/zÚ³~ï¿½ï¿½aï¿½ï¿½ï¿½&ï¿½ï¿½ï¿½ÂEï¿½ï¿½>ï¿½ï¿½ï¿½Mï¿½'6f".z #Vz{"ï¿½ftQ"ï¿½ ï¿½7jï¿½vï¿½F&ï¿½ï¿½>sï¿½ï¿½ï¿½g~6L(r)Uï¿½0ï¿½ï¿½ï¿½ï¿½aï¿½lï¿½ï¿½aï¿½ï¿½qXï¿½3/4ï¿½LÙ b%Mï¿½(;"ï¿½ï¿½{ï¿½aï¿½_(c)=Vï¿½Ø 8F ï¿½pï¿½Jooï¿½ï¿½ï¿½3/4 ï¿½o ï¿½"`3/4a[ï¿½-ï¿½ï¿½)ï¿½gHU3ï¿½(r):ï¿½ï¿½#vw%Hï¿½ï¿½ï¿½"Ú^;ï¿½ï¿½D(r)ï¿½...,ï¿½ï¿½Lï¿½ï¿½-yÑï¿½ï¿½4ï¿½ï¿½ #Tï¿½ &ï¿½ï¿½ï¿½ï¿½ï¿½x5ï¿½Kï¿½qdï¿½ï¿½ï¿½ï¿½1/2^ï¿½Ngï¿½r:ï¿½O"Wï¿½rï¿½óºï¿½ï¿½hï¿½"ï¿½Ezï¿½Fï¿½5&ï¿½-(c)ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½NDc5ï¿½h5(r) BSï¿½ï¿½C_ï¿½7"q(r)xï¿½jï¿½jï¿½5?J u]Wï¿½(r)wï¿½ï¿½;,ï¿½kï¿½"Ô¶ï¿½_Ö¶ï¿½.fYï¿½F-ï¿½ï¿½/ï¿½+ï¿½ï¿½ï¿½3/4ï¿½ï¿½ï¿½Qï¿½2fï¿½ï¿½ï¿½ï¿½Rï¿½Fï¿½ ï¿½ï¿½ï¿½ï¿½ï¿½~ï¿½;*-:ï¿½j?ï¿½ï¿½ï¿½Tï¿½yï¿½ï¿½mï¿½ï¿½ï¿½ï¿½Tï¿½A(ï¿½ï¿½gï¿½jBï¿½3ï¿½1/2Siï¿½1/2ï¿½ï¿½ï¿½fï¿½ï¿½ï¿½ ï¿½ï¿½ï¿½I`<(c)Dï¿½ï¿½{ï¿½~ï¿½ï¿½#ï¿½ ï¿½"qï¿½Ø]/tï¿½ï¿½ï¿½vï¿½ï¿½ï¿½ï¿½Û±eï¿½ï¿½ï¿½Oï¿½|ï¿½Kï¿½ï¿½*ï¿½Kï¿½fï¿½ï¿½oï¿½;ï¿½Íºï¿½ï¿½ï¿½ocskï¿½×¯ï¿½ï¿½/ï¿½^ï¿½Vï¿½Hï¨ dï¿½ï¿½ï¿½nVï¿½ï¿½-ï¿½1/2Veï¿½ï¿½ï¿½Mï¿½ï¿½ ï¿½Lï¿½ï¿½-ï¿½ï¿½A41/4I(c)4ï¿½3/4ï¿½ï¿½ï¿½oï¿½|Vï¿½ï¿½Eï¿½ï¿½'ï¿½ï¿½ï¿½ï¿½nc#~ï¿½rï¿½ï¿½ï¿½ï¿½ï¿½ Aï¿½ ï¿½xo1/4Mï¿½ï¿½ï¿½"(r)ï¿½'ï¿½-.ï¿½ï¿½ ï¿½uH-AHÏ¯Azï¿½g>r ï¿½VGï¿½r\ï¿½Wï¿½iï¿½\<*)ï¿½ï¿½ ï¿½1/4ï¿½Hï¿½ï¿½ï¿½ï¿½kX!mKï¿½ï¿½7...Pï¿½vSï¿½6\ï¿½ï¿½ï¿½ï¿½iï¿½È¦ï¿½ca"(tm)(c)ï¿½,3/4 ï¿½ï¿½-ï¿½ï¿½f5ï¿½ï¿½ É¦-ï¿½&ï¿½uï¿½ï¿½ ï¿½>{ï¿½zï¿½>XR2ï¿½ï¿½e(uï¿½ Ù ï¿½ï¿½Dï¿½ï¿½Sï¿½Ï"ï¿½ï¿½ Kï¿½-"ï¿½'ï¿½p&ï¿½Gï¿½r ï¿½'2ï¿½Fï¿½ï¿½ï¿½JGï¿½ï¿½fï¿½ï¿½'ï¿½%åï¿½ï¿½+cIY)ï¿½'-ï¿½ï¿½ï¿½ï¿½liï¿½!ï¿½|ï¿½Yï¿½\^yï¿½zg×³×°[ï¿½1/4 1Ó8ï¿½ï¿½hï¿½ï¿½ï¿½ï¿½}ï¿½Ý¥pzpP ï¿½4ï¿½ ï¿½m ×ï¿½ï¿½|ï¿½o?ï¿½h-ï¿½#ï¿½Luï¿½ /ï¿½1/2#Wï¿½ï¿½ï¿½h"( !6;"ï¿½ï¿½ mï¿½0ZÄ¥ï¿½<ï¿½3/4ï¿½ "'"ï¿½tï¿½Pï¿½ï¿½>uï¿½ï¿½ï¿½D2 rb&qï¿½Øï¿½>ï¿½ï¿½{"7ÇOï¿½< ï¿½Hï¿½ï¿½xCï¿½ï¿½ï¿½O3/4;ï¿½ï¿½... sï¿½ï¿½ Og(tm)P"ï¿½"=9ï¿½3ï¿½ï¿½9S-ï¿½?ï¿½%8ï¿½!ï¿½ï¿½ï¿½6ï¿½&-ï¿½+Iï¿½ï¿½8Þ· -)e)f'X5rï¿½-(tm)ï¿½[IHï¿½9ï¿½-#ï¿½Pï¿½ï¿½z(ï¿½ï¿½8p3J[a+...ï¿½cpRzWï¿½96ã¬´ï¿½ï¿½ï¿½(tm)ï¿½hLaï¿½'36ï¿½"Fï¿½ï¿½-u #xï¿½(8e 'ï¿½Gï¿½ï¿½dï¿½ï¿½ï¿½Nzjï¿½&Yï¿½]ï¿½Êï¿½K9'rï¿½ï¿½e(r)ï¿½I*b_ï¿½Tï¿½jï¿½:"arLï¿½D[ ï¿½Iï¿½G(P&ï¿½>-ï¿½f:i'S3ï¿½ï¿½ï¿½ï¿½Wï¿½,ï¿½_Gï¿½qï¿½9ï¿½<e'-ï¿½ ï¿½|}zuvï¿½Oï¿½(r)ï¿½@ï¿½-ï¿½SU4ï¿½6Vï¿½nï¿½ï¿½Qï¿½ï¿½>Mï¿½,xSï¿½ï¿½&- ï¿½eï¿½6k1/4ï¿½qï¿½ï¿½Iï¿½E,ï¿½}@ï¿½tÙ£ï¿½ï¿½ ï¿½_`8ï¿½t "+ï¿½<Qï¿½ï¿½(7xï¿½zï¿½R/>ï¿½{¬w"G(vï¿½ï¿½Ç¿1/2?ï¿½ï¿½Ojï¿½zBï¿½fjï¿½,Yï¿½8ï¿½ï¿½(r)ï¿½ï¿½gï¿½ (c)ï¿½!ï¿½"ï¿½ï¿½ï¿½H"ï¿½ï¿½"!1/4Zï¿½ï¿½)v[ï¿½Cvï¿½'ï¿½ï¿½ï¿½ï¿½hï¿½Hï¿½ï¿½ "s\$e\ï¿½-ï¿½ï¿½@dï¿½ï¿½ï¿½ï¿½Dï¿½ßï¿½rï¿½ï¿½ï¿½ ï¿½~ï¿½ï¿½l2vhï¿½<l'ï¿½ï¿½" ï¿½ 3/4ï¿½ï¿½ï¿½ï¿½\ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ ï¿½wï¿½ï¿½ï¿½Jï¿½ï¿½O \ï¿½ï¿½ï¿½Lï¿½^ï¿½ to-2Ï7ï¿½:ï¿½Kï¿½1/2[Sï¿½ï¿½ï¿½ï¿½ï¿½A[(ï¿½"S3/4ï¿½rï¿½ï¿½? 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• Word count: 5020
4. Two in a line

So there are 3 x 3 = 9 ways diagonally. (in this direction) For the diagonal patterns there are two different directions. 1. The direction in the squares above 2. The direction in the squares below This means that if there are 9 in one direction then there are 2 x 9 = 18 in both directions Therefore there are 12 (horizontally) + 12 (vertically) + 18 (diagonally) = 42 EXTEND INVESTIGATION Based on the idea above we can investigate other size grids. We could start with square grids but it is just as easy to look at rectangular grids.

• Word count: 1465
5. How does Nike benefit from competitive advantage?

of letters Working Arrangement 1 1 1 2 2x1 2 3 3x2x1 6 From this initial assessment the rule for any size word combination, where all the letters are different, is a= n! where n is the number of letters in the word and a is the total number of arrangements. I would predict, therefore, that there will be 4! or 24 arrangements in the word Lucy. My check is below:- 1) LUCY 7) ULCY 13) CULY 19) YLUC 2) LUYC 8) ULYC 14) CUYL 20) YLCU 3) LYCU 9) UCYL 15) CYUL 21) YUCL 4) LYUC 10) UCLY 16)

• Word count: 2491
6. Emma's Dilemma

that do not have any repeats and are one, two, three, four and five letters long: A JO OJ MAX MXA AMX AXM XMA XAM MIKE MIEK MKIE MKEI MEIK MEKI IMKE IMEK IKME IKEM IEMK IEKM KMIE KMEI KIME KIEM KEMI KEIM EMIK EMKI EIMK EIKM EKMI EKIM MINTY MNITY MTINY MYINT MINYT MNIYT MTIYN MYITN MITNY MNTIY MTNIY MYNIT MITYN MNTYI MTNYI MYNTI MIYNT MNYIT MTYIN MYTIN MIYTN MNYTI MTYNI MYTNI IMNTY INMTY ITMNY IYMNT IMNYT INMYT ITMYN IYMTN IMTNY INTMY ITNMY IYNMT IMTYN INTYM ITNYM IYNTM IMYNT INYMT ITYMN IYTMN IMYTN INYTM ITYNM IYTNM NMITY NIMTY NTMIY

• Word count: 800
7. Emma's Dilemma

Here is another example using the name TIM. Total number of letters: 3 Previous number of combinations: 2 3x2=6 Here is another example using the name DAISY Total number of letters: 5 Previous number of combinations: 24 24x5=120 After doing this I realised that I can work out the amount of combinations using factorial notation. This is when a number is multiplied by the previous consecutive numbers. For example, using the letter 6 you would do this: 6x5x4x3x2x1 I realised this is because if I could find the total number of combinations by multiplying the total number of letters by the previous number of combinations it was the same as multiplying the total number letters by its previous consecutive numbers.

• Word count: 812
8. Emma's Dilemma.

ANKIR AKRNI ARKIN AIKRN ANIKR AKRNI ARKNI AIKNR ANIRK AKINR ARNKI AIRKN ANRKI AKNIR ARNIK AIRNK ANRIK AKNRI RINKA RNKAI RKAIN RAKIN RINAK RNKIA RKANI RAKNI RIKAN RNAKI RKINA RANIK RIKNA RNAIK RKIAN RANKI RIAKN RNIAK RKNIA RAIKN RIANK RNIKA RKNAI RAINK NIKAR NKIAR NAIRK NRAKI NIKRA NKIRA NAIKR NRAIK NIARK NKARI NARIK NRKAI NIAKR NKAIR NARKI NRKIA NIRAK NKRAI NAKIR NRIKA NIRKA NKRIA NAKRI NRIAK INKAR IKARN IARNK IRNKA INKRA IKANR IARKN IRNAK INARK IKRNA IANKR IRKNA INAKR IKRAN IANRK IRKAN INRAK IKNRA IAKRN IRANK INRKA IKNAR IAKNR IRAKN For this, there are 120 possible combinations RESULTS Table of

• Word count: 1590
9. Emma's dilemma

First of all we have got 4 possibilities (4 letters) and then 3 possibilities(by leaving first letter "L") and then 2 possibilities 9by leaving first two letters "L"U") after that one possibility which is the last letter . Therefore 4*3*2*1=4! Which is equal to 24, where as we can see this from tree diagram (which is shown by using different colours) > Part 2 I am going to investigate the number of different arrangements of the letters of EMMA'S name. EMMA 1. EMMA 2. EMAM 3. EAMM 4. AMEM 5.

• Word count: 2463
10. Emma's Dilemma

Then, if I re-arrange a three-letter word there will be more re-arrangements due to the larger number of letters. > I will start by finding out how many times 'Lucy' can be re-arranged. Below are all the possibilities of re-arranging the name 'Lucy.' LUCY UCYL CYLU YLUC LUYC UCLY CYUL YLCU LYUC ULCY CLYU YULC LYCU ULYC CLUY YUCL LCUY UYCL CULY YCLU LCYU UYLC CUYL YCUL There are 24 ways in which I could re-arrange the name 'Lucy.' I am now going to find out how there are 24 different ways in which I can re-arrange 'Lucy.' Below is a table of how many times you can re-arrange groups of letters which have less letters than in 'Lucy.'

• Word count: 1205
11. Emma Dilemma

For example I already know that Lucy has 6 combinations with L at the front if I multiply that by the number of letters in the name which is 4 I should get the answer. 6x4=24. 24 is the correct answer, so there is 24 different ways (combinations) of writing Lucy. This method will not work with a name with 2+ letters the same only with a name with its entire letters different. To work out a name with 2 letters the same I will need to some more work, which is my next step.

• Word count: 1785
12. Emma's Dilemma

gives 24. This means that all names with 4 different letters will have a total of 24 different combinations. I realized that the total number of combinations is the number of letters multiplied by the previous number of combinations. Name: Ian Total number of letters: 3 Previous number of combinations: 2 3 X 2 = 6 Summary of Parts 1 and 2: EMMA and LUCY both have the same number of letters in their names, however LUCY has twice as many different letter combinations.

• Word count: 2695
13. Emma's Dilemma

So the calculations for this would be 4 x 6 = 24. For a five-lettered word there are 24 different arrangements with each letter at the beginning. So 5 x 24 =120. So I predict that a five lettered word, which has no repeated letters in it will have a total of 120 different arrangements. Five-lettered word KEANO KAENO KANEO KANOE KAEON KAOEN KAONE KEAON KEOAN KEONA KENOA KENAO KNEAO KNEOA KNAEO KNAOE KNOAE KNOEA KOEAN KOENA KONEA KONAE KOANE KOAEN EKNAO EKNOA EKONA EKOAN EKAON EKANO ENKAO ENKOA ENOKA ENOAK ENAOK ENAKO EANKO EANOK EAKNO EAKON EAONK EAOKN EOAKN

• Word count: 2900
14. Investigate the number of different arrangements of letters in different words

x 2 = 6, the number of different arrangements. In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4! which is called 4 factorial. I found this out in the school library which is the same as 4 x 3 x 2 but times 1 but this make no difference to the result. This means that the formula for the number of arrangements for a word with no repeated letters is: n!

• Word count: 1192
15. Emmas dilemma.

Now I will investigate a 2 letter word. IT TI There are two possible arrangements, one for each letter. I will now draw a table of results: Number of letters in word Number of arrangements 1 1 2 2 3 6 4 24 5 ? From this table I can work out the number of combinations for a four letter word, without knowing what it is, I can multiply the previous number of arrangements by the present number of letters in the word, this also works for a 3 letter word and so on.

• Word count: 1493
16. Emma's dilemma.

If you were to colour code the Ms in EMMA then you may have 24 combinations. I am going to investigate this. * EMMA * EMMA* * EMAM * EMAM* * EAMM * EAMM* * MEMA * MEMA* * MAME * MAME* * MMAE * MMAE* * MMEA * MMEA* * MEAM * MEAM* * MAEM * MAEM* * AMME * AMME* * AMEM * AMEM* * AEMM * AEMM* All of the combinations with a * next to them are cancelled out because if they were not colour coded then they would look exactly the same as the combination above it.

• Word count: 759
17. I am investigating the number of different arrangements of letters in a word.

=24 + * E (digo) =24 + * G (dieo) =24 + * O (dieg) =24 + =120 Total no. Of letters No repeated letters 1 1 2 2 3 6 4 24 5 120 In a 4-letter name such as Lucy, the way to work out the number of combinations is to do 4 x 3 x 2 x 1 or 4 factorial (4!). . Factorial is a number multiplied by the previous consecutive numbers. Factorial notation is symbolised using an exclamation mark! So I came to conclude that the formula to work out any the combinations is the numbers of letters factorial equals arrangements (l!

• Word count: 787
18. Emma&#146;s Dilemma.

Combinations: 120 From the above I can work out a table of results to enable me to work link the number of letters all different with its amount of combinations. No. of letters, all different 2 3 4 5 6 7 8 9 10 Combination 2 6 24 120 ? ? ? ? ? From the above table I can recognise that the number of combinations ina word where all the letters are different is the total number of letters factorial.

• Word count: 1575
19. Emma's Dilemma.

lucy 2) luyc 3) lcuy 4) lycu 5) lyuc 6) lcyu 7) cyul 8) clyu 9) cylu 10) cluy 11) culy 12) yulc 13) cuyl 14) ulyc 15) ucyl 16) ulcy 17) ucly 18) uycl 19) uylc 20) ylcu 21) ycul 22) yclu 23) yucl 24) yluc There are 24 different possibilities in this arrangement of 4 letters all different. Double the amount as before with Emma's name, which has 4 letters and 3 different. I have noticed that with Lucy there are 6 possibilities beginning with each different letter.

• Word count: 2812
20. Dave's Dillemma.

EMMM has 4 arrangements because 3 of the letters are the same. MMMM has 1 arrangement because all of the letters are the same. It is clear from these results that as the number of letters which are different increases, the number of arrangements decreases. There is an obvious pattern. The number of arrangements for the previous arrangement gets divided by the number of letters that are the same in that particular name. e.g. Arrangements for EMMA = 24 = 12 2 Arrangements for EMMM = 12 = 4 3 Arrangements for MMMM = 4 = 1 4 Also, if 4!

• Word count: 1728
21. Emma&#146;s Dilemma.

Wrote down the 2 other possibilities. There were six possibilities for the word beginning with L. I repeated the process with the letters U, Y and C. I found out that there were 6 possibilities for each of these letters and then I multiplied 6 by 4 (the number of letters) which gave me 24 possibilities for the word LUCY. Part 2 Introduction In this investigation, I am going to find out the different arrangements of the letters of the name EMMA Here are the possibilities for the name EMMA EMMA EMAM EAMM MEMA MEAM MMEA MMAE MAEM MAME AMME

• Word count: 772
22. Emma's Dilemma.

or I could work out all the other arrangements by using the branching method on all the other letters in the name. Results: Firstly I am going to investigate names with different number of letters. I will investigate these names; Jo Sam Lucy Mandy After I have investigated these names I will have enough information to predict how many arrangements of a six-letter name there will be. Jo J o Jo The name Jo only has two different arrangements. O j Oj Sam a m Sam S m a Sma s m Asm A m s Ams s a Msa

• Word count: 1848
23. Emma's Dilemma.

x 2 = 6, the number of different arrangements. In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4!, which is called 4 factorial which is the same as 4 x 3 x 2. So, by using factorial (!) I can predict that there will be 40320 different arrangements for an 8 letter word. The formula for this is: n! = a Where n = the number of letters in the word and a = the number of different arrangements.

• Word count: 1993
24. Emma's Dilemma.

KMAR 22. KMRA 23. KARM 24. KAMR As you can see the name MARK also has 24 different combinations. Calculations The number of combinations for the name Lucy can be worked out in the following ways. 1. no. of letters + 2 x no. of letters [4+2 x 4 = 24] 2. no. of letters factorial [4! = 1x2x3x4 = 24] To see if the above methods work, we can test them on the three (different) letter name Ian. Below are the different combinations of the name Ian. 1. IAN 2. INA 3. ANI 4. AIN 5. NIA 6. NAI We can see that the name Ian has 6 possible combinations. Now I will test the methods mentioned previously.

• Word count: 1262
25. Emma's Dilemma.

I already know the number of arrangements for a four -lettered name 'LUCY' that is 24 arrangements. This is quadruple the number of arrangements than 'JIM'. Now I shall gather my results and form a table so I can try to predict the number of arrangements in a 5 lettered word with all different letters. N R A NAL 2=JO NO 2 1 3=JIM NO 6 2 4=LUCY NO 24 6 NAL= Number of arrangements for using a letter in the beginning of the word e.g. for LUCY, putting L in the beginning of the word and seeing how many arrangements there are.

• Word count: 3131