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• Level: GCSE
• Subject: Maths
• Word count: 1149

# Emma's Dilemma

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Introduction

GCSE COURSEWORK Matthew Kilgour 25/8/01 Arrangements for Emma: 4 letters, 2 the same. 1.emma 2.emam 3.eamm 4.mmae 5.mmea 6.meam 7.mema 8.mame 9.maem 10.amme 11.amem 12.aemm In a word/name with 4 letters and none the same, then there are 24 possible arrangements. There are 12 possibilities; note that there are 4 total letters and 3 different. Emma has four letters, three of which are different. This means that it only has 12 possible combinations. I had noticed that 4 (number of total letters) multiplied by 3 (the number of different letters) equals 12. What if all the letters were different like in Lucy? Arrangements for Lucy: 4 letters, all different, none the same. As I stated before, in a word with 4 letters and none the same, there are 24 possible combinations, double the number of combinations for Emma, which has 4 letters and 3 the same. 1.lucy 2.luyc 3.lcyu 4.lycu 5.lcyu 6.lyuc 7.ulcy 8.ucly. 9.uylc 10. ...read more.

Middle

there were 24 combinations, and with 5 letters all that is happening is that an extra point to start form is added. Eg 24 combinations a 4 letter word, with an extra in front of it. So 120=24 5 Previously in the 4-letter word, 24 divided by 4 equals 6, the number of possibilities there were for each letter. To make things simpler and east to under stand I shall use the name Xlucy. 1.xlucy 2.xucyl 3.xcylu 4.xycul 5.xluyc 6.xucly 7.xcyul 8.xyclu 9.xlcuy 10.xulcy 11.xculy 12.xyulc 13.xcyu 14.xulyc 15.xcuyl 16.xyucl 17.xlyuc 18.xuycl 19.xclyu 20.xylcu 21.xlycu 22.xuylc 23.xcluy 24.xyluc I have done the first 24 in this sequence, as it makes it easy to see that to obtain the other combinations all that is needed is to replace the first letter with one of the others, so I will not list all 120 possibilities. I can see that the numbers of possibilities for different arrangements are going to increase as more letters are used, as long as the letters are different. ...read more.

Conclusion

4 factorial is 24, but I could only find 12 possibilities, this means that factorising will not work alone. There for if there are repeated letters, you answer must be divided by the number of repeated letters, eg Emma, 4!= 24 24=12 the number of 2 combinations . It is easier to understand in terms of x And y. I will start with xxyy: Arrangements for xxyy: 4 letters, 2 pairs 1.xxyy 2.xyxy 3.xyyx 4.yxxy 5.yxyx 6.yyxx There are 6 arrangements. Arragements for xxxyy: 5 letters, a pair and 3 the same 1.xxxyy 2.xxyxy 3.xxyyx 4.xyxyx 5.xyxxy 6.xyyxx 7.yyxxx 8.yxxxy 9.yxyxx 10.yxxyx There are 10 different arrangements . arrangement for xxxxy: 1.xxxxy 2.xxxyx 3.xxyxx 4.xyxxx 5.yxxxx There are 5 different arrangements for this instance. If I use xxxyy as an example, there are 2y's and 3 x's in a total of 5 unknowns. The best formula I could come up with is the number of total letters factorial, divided by the number of x's, y's etcfactorised and multiplied. eg A five letter word like xxxyy; this has 3 x's and 2 y's. 1x2x3x4x5 = 120 / 12 = 10 1x2x3 x 1x2 ...read more.

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