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  • Level: GCSE
  • Subject: Maths
  • Word count: 1333

Emma's dilemma.

Extracts from this document...

Introduction

Bhavika Patel  11DB

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Emma is playing with an arrangement of the letters in her name.

  • One arrangement is Emma
  • A different arrangement is Meam
  • Another arrangement is Aemm
  • Investigate the number of different arrangements for Emma’s name.
  • Emma has a friend called Lucy.  Investigate the number of different arrangements for Lucy’s name.
  • Choose some different names and investigate the number of different arrangements for the names.

A number of X’s and Y’s are written in a row such as:

                xxxx…….xx…..yyy….xyyyy

  • Investigate the number of different arrangements for the row of letters.

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

The number of different arrangements for Emma’s name

  • EMMA        MEAM        MMEM        MAEM        
  • EMAM        AEMM        MMAE        MEMA = 12 arrangements
  • EAMM        AMEM        MAME        AMME

The number of different arrangements for Lucy’s name

  • LUCY        LYCU        UCYL        CYLU        CULY        
  • LUYC        LYUC        UCLY        CYUL        CUYL
  • LCYU        ULCY        UYCL        CLUY        YLUC
  • LCUY        ULYC        UYLC        CLYU        YLCU
  • YUCL        YULC        YCLU        YCUL  = 24 arrangements

For one letter there will obviously be only one combination

E.g.   B = one arrangement

For two letters there will be two arrangements:

E.g.          

  • TI and IT = two arrangements

For three letters there will six arrangements;

...read more.

Middle

CANIE        CINEA        CNAEI        CEAIN        CEINACAENI        CINAE        CNEAI        CEANI     …………..

The above arrangements show the first 24 arrangements for the five-letter name CAINE.  The first 24 arrangements all

Have the letter C in the front and in this same way if I was to carry on every other letter would come to the front and therefore there would be 24 x 5 arrangements, which is 120.

        If two of the letters in a name are the same then there cannot be a name with just one letter but if it has two letters then there is simply one arrangement.

If a name has three letters two of which are the same then there will be two arrangements as shown below:

  • LEE                EEL                ELE

If a name has four letters with two letters the same then it will have 12 arrangements (i.e. Emma) and if it has 5 letters of which two are the same then it will have 60 different arrangements:

  • GEMMA                GAMME                GEMAM
  • GAMEM                GMMEA                GMMAE
  • GMEAM                GMAEM                GMEMA
  • GMAME                GAEMM                GEAMM       ………

These are the first 12 arrangements of the name Gemma that has two letters the same.

...read more.

Conclusion

        e.g. Bhavika has seven letters two of which are the same (a). This name would therefore have 2520 arrangements because 7!/ 2! = 2520

        We could work out this number using the initial method and to do that we would multiply 360 by 7, which equals 2520.

        Another sort of name (more a word) that I investigated was SENSES.  This is a good word because it has one letter that is repeated three times and another letter that is repeated two times.

        The formulae to figure the number of arrangements for this is 6!/ 3!2! (3! For the three S’s and 2!for the two E’s and 6! For the total number of letters), that gives the answer 60. We can check this by writing the arrangements out:

  • SENSES                NSSSEE                NSSEES
  • NSSESE                NSEESS                NEESSS
  • NESESS                NESSSE                NESSES
  • NSESSE                NSESES

It is visible that the letter N comes in front ten times therefore the other 5 letters will come forward 10 times each and there will then be 60 combinations.

        For the X’s and y’s the same method then applies:

Total number of letters!/number of times letter repeated! X number of times letter repeated!

For example:    xxxxxyyyyyyy = 12! / 5!7! =792 arrangements

There is also an x!  Button on most scientific calculators and this also helps a great deal!!!!!

...read more.

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