Emma is playing with arrangements of the letters of her name to see how many combinations there are.

EMILY                EILYM                ELMYI                EYMLI

EMIYL                EILMY                ELMIY                EYMIL

EMYLI                EIMYL                ELIYM                EYILM

EMYIL                EIMLY                ELIMY                EYIML

EMLYI                EIYML                ELYMI                EYLMI

EMLIY                EIYLM                ELYIM                EYLIM

I have now done all the combinations for Emily beginning with E and found 24 different arrangements. I predict that there will be 120 different combinations in total, a further 24 combinations for each of the remaining letters: M, I, L and Y. I will continue with my method to see if my prediction is correct.

MILYE                MLYEI                 MYEIL                 MEILY

MILEY                MLYIE                 MYELI                 MEIYL

MIYEL                MLIYE                 MYLIE                 MELIY

MIYLE                MLIEY                 MYLEI                 MELYI

MIEYL                MLEYI                 MYIEL                 MEYLI

MIELY                MLEIY                 MYILE                 MEYIL

I have found a further 24 arrangements beginning with M.

ILYEM                IYEML                 IEMLY                 IMLYE

ILYME                IYELM                 IEMYL                 IMLEY

ILEMY                IYMLE                 IEYLM                 IMYEL

ILEYM                IYMEL                 IEYML                 IMYLE

ILMYE                IYLEM                 IELYM                 IMELY

ILMEY                IYLME                 IELMY                 IMEYL

LYEMI                LEMIL                 LMIYE                 LIYEM

LYEIM                LEMLI                 LMIEY                 LIYME

LYMIE                LEIYM                 LMYEI                 LIEMY

LYMEI                LEIMY                 LMYIE                 LIEYM

LYIEM                LEYMI                 LMEIY                 LIMYE

LYIME                LEYIM                 LMEYI                 LIMEY

YEMIL                YMILE                 YILEM                 YLEMI

YEMLI                YMIEL                 YILME                 YLEIM

YEILM                YMLEI                 YIEML                 YLMIE

YEIML                YMLIE                 YIELM                 YLMEI

YELMI                YMEIL                 YIMLE                 YLIEM

YELIM                YMELI                 YIMEL                 YLIME

My prediction was correct – there are 120 different combinations for a five-letter name with no repeated letters. There were 24 combinations for each of the five letters in the name. So, 5 x 24 = 120.

I will now try a name with a repeated letter. I will use the name Kerry.

KERRY                ERRYK                 RYERK                 RRKYE

KERYR                ERRKY                 RYEKR                 RRKEY

KEYRR                ERYRK                 RYRKE                 RREKY

KYRRE                ERYKR                 RYREK                 RREYK

KYRER                ERKYR                 RYKRE                 RRYKE

KYERR                ERKRY                 RYKER                 RRYEK

KRERY                EYRRK                 RERKY                 RKERY

KRREY                EYRKR                 RERYK                 RKEYR

KRRYE                EYKRR                 REYKR                 RKYRE

KREYR                EKRRY                 REYRK                 RKYER

KRYRE                EKRYR                 REKRY                 RKREY

KRYER                EKYRR                 REKYR                 RKRYE

YKERR                YERRK                 YRREK                 YREKR

YKRER                YERKR                 YRRKE                 YRKRE

YKRRE                YEKRR                 YRKER                 YRERK

There are 60 combinations for a five-letter name with one repeated letter.

I will now try a name with two repeated letters I will use the name Hanna.                  

HANNA                              ANNAH                          NANAH                                                          

HANAN                              ANNHA                          NANHA

HAANN                              ANHNA                          NAHNA

HNNAA                              ANHAN                          NAHAN

HNANA                              ANANH                          NAAHN

HNAAN                              ANAHN                          NAANH

                                             AAHNN                          NNAAH

                                             AANHN                          NNAHA

                                             AANNH                          NNHAA

                                             AHNNA                          NHAAN

                                             AHNAN                          NHANA

                                             AHANN                          NHNAA

I have found 30 different combinations for a five-letter name with two repeats.

I can see from my table that the pattern for the results for five-letter names is the same. The number of arrangements is halved each time a letter is repeated.

I am trying to find a formula to find the number of combinations for any given number of letters. I will find results for six-letter names to help me find the formula. I predict that there will be 720 combinations for a six-letter name with no repeats. I think this because there are 6 combinations beginning with NIC;

NICOLA                          NICALO                           NICLOA

NICOAL                           NICAOL                          NICLAO

Therefore there will be 6 combinations each for NIL, NIA and NIO.

Eg. NILCAO                      NILACO                        NILOCA

      NILCOA                      NILAOC                         NILOAC etc.

There will therefore be 24 combinations for NI, 24 for NC and so on. Altogether there will be 120 combinations beginning with the letter N. This means that there will be 120 combinations for each of the six letters.

So, 6 x 120 = 720

My prediction was correct.

To work out the formula I will use a table displaying my results for names with no repeats.

             

                                       x2         x3           x4            x5           x6

I can see from my table that I have to multiply ‘c’ each time by ‘n+1’.

So, to find the number of combinations for a seven-letter name with no repeated letters, I could do   720 x 7 = 5040

 I could also do  1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040

I can now find out the number of combinations for any name with no repeated letters.

Eg. For 8

1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320

I could also use the factorial key that I have been shown on the calculator. For example, to find the number of combinations for a seven-letter name with no repeats on a calculator:

7     2nd F     n!      =   5040

I would write this as 7! = 5040

Now that I have found out the sum to find factorials with no repeats, I will go back to my earlier investigation into names with repeated letters.

So far I have found the following information:

1 Repeat:

                                                    x3         x4         x5         x6

I can see that the pattern is the same as for no repeats except that instead of    1 x 2 x 3 etc. it is  1 x 3 x 4 etc.  with the two missed out. I can write this as  7!

            2

Now I will try names with 2 repeated letters.

2 Repeats:

                                                       x5        x6         x7

I can write this as  n!

                               4

So,

• The sum for no repeats: n!
• The sum for 1 repeat:    n!

                                            2

• The sum for 2 repeats:  n!

                                           4

                                                                                                 

Now that I have finished this part of my investigation, I will go onto the next part. I will investigate the number of different arrangements for any word with any number of letters. I will use the letters X and Y to start with.

X                              I can see that for these, the number of

XY                           combinations is the same as the number of

XXY                        letters. I can see this as for XXXY there are

XXXY                     possible positions for the Y to be in and, as all

XXXXY                  the other letters are the same, 4 possible

XXXXXY               combinations.       XXXY

XXXXXXY                                         XXYX

                                                              XYXX

                                                              YXXX

I will now draw up a table to show my results more clearly.

I will now try 3 different letters.

XYZ                  There are six different combinations for 3 different

XZY                   letters. It is the same as my previous result for a

YZX                   three-letter word.

YXZ

ZYX

ZXY

I will now add another X to see hoe many combinations there are.

XXYZ               XYZX              YXXZ             ZXXY

XXZY               XZXY              YXZX             ZXYX

XYXZ               XZYX               YZXX            ZYXX

I found 12 combinations.

I will now compare the results for both parts of my investigation.

I can see from my table of results that x (number of arrangements for 1x and n y) multiplied by the previous number c (number of combinations) equals the next c.

Eg.

      C                  2          6          24   

      x                   2          3           4

                                           4 × 6 = 24

I can write this as nfactorial divided by xfactorial (x standing for the number of arrangements for 1x and ny).

Or

n!

×!

eg.  

         

              2 ÷ 2 = 1

              6 ÷ 3 = 2

(c)        24 ÷ 4 = 6                previous

          120 ÷ 5 = 24                    (c)

          720 ÷ 6 = 120

I am now trying to find out a formula to find out the number of combinations for any word with any number of repeated letters. I will put all my results into a table to show them more clearly.

I know that  n!  equals the number of combinations for any word with no  

                    x!

repeated letters and  n!  is for one repeated letter etc. However, I want to

                                 2!

find out the formula for working out the number of combinations for words with letters repeated more than once.

Eg.   XXXYYY

The formula for this could be written as:

        6!           (no. of letters)

                                              3! × 3!                 (no. of x’s > y’s) 

Or

          720    = 20

36

I will write out all the possible combinations for XXXYYY to check my result.

XXXYYY

XXYXYY

XXYYXY

XXYYYX

XYXXYY

XYXYXY

XYXYYX

XYYXXY

XYYXYX

XYYYXX

YXXXYY

YXXYXY

YXYXXY

YXYXYX

YXYYXX

YYXXXY

YYXXYX

YYXYXX

YYYXXX

YXXYYX

My formula worked correctly – I have found 20 possible combinations for XXXYYY.

I can now do a formula for any word with any number of letters and any number of repeats.

For example:

                         For XXYYZZ it would be

6!    

2! 2! 2!

For XXYZABC it would be

7!

2! 1! 1! 1! 1! 1!

My overall formula would be:

n!

a!b!c!d!e!f!g!h!i!j!k!l!m!n!o!p!q!r!s!t!u!v!w!x!y!z!

This can be used to find out the number of combinations for the letters of any word. I would just have to substitute the number of each letter for the letter. So, instead of a!, I would put in the number of times that a appears in the word I was trying to find out the number of combinations for.

For example:

‘Maths’ would be:

5!

1! 1! 1! 1! 1!

Which would equal 120 different combinations.

Whereas ‘classes’ would be:

7!

1! 1! 1! 3! 1!

Which would equal 840 different combinations.