I have noticed that as the number of letters in a word increases, a increases.
a, also increases dramatically, a for when n is 3, is triple the amount of a when n is 2. When n is 4 a is quadruple the amount of arrangements for when n is 3. From this I might be able to predict that a for when n is 5 is 5 times the amount of a for a 4-lettered word.
As I found a for when n is 2, which is 2 arrangements. So when n is 3 there shall be 2 arrangements for each letter in the beginning. 3*2=6 Therefore when n is 4 there are 6 arrangements for each of the letter in the beginning 6*4=24
From these observations I should be able to predict a when n is 5 having different letters. I shall use the name Ralph.
Prediction for Ralph
I can predict a for the name Ralph by looking at my observations from the table above. As a when n is 3 is triple the amount of arrangements of when n is 2. So when n is 4 the number of arrangements is quadruple the amount of arrangements for when n is 3. Therefore a for when n is 5, is 5 times a for when n is 4. So because there were 24 arrangements when n is 4, 5*24=120. So I predict that there will be 120 arrangements for when n is 5.
The other reason why I predict this is secondly because the number of arrangements for when n is 2 is 2 arrangements. So when n is 3 there shall be 2 arrangements for each letter in the beginning. 3*2=6 so when n is 4 there are 6 arrangements for each of the letters in the beginning 6*4=24 therefore when n is 5 there are 24 arrangements for each of the letters in the beginning 24*5=120
ARRNAGEMENTS FOR RALPH
I want to show my results in a systematic manner so I will show a having R in the beginning of the name. I should from my prediction find 24 arrangements with R in the beginning.
1)RALPH 2)RALHP 3) RAHLP 4)RAHPL 5)RAPHL 6)RAPLH 7)RPALH 8)RPAHL 9)RPHLA 10)RPHAL 11)RPLAH 12) RPLHA 13)RHALP 14)RHAPL 15) RHLPA 16) RHLAP 17)RHPLA 18) RHPAL 19)RLPAH 20) RLPHA 21)RLAHP 22)RLAPH 23)RLHPA 24)RLHAP
My prediction was right as there would be 24 arrangements for each of the letters in the beginning of this word this is shown because there are 24 arrangements with R in the beginning of the word, so therefore there should also be (24 arrangements for A) (24 arrangements for L) (24 arrangements for P) (24 arrangements for H) I shall make sure that this is true by investigating the rest of the arrangements for RAPLH
25)ARLPH 26)ARLHP 27)ARPHL 28)ARPLH 29)ARHPL 30)ARHLP 31)ALPRH 32)ALPHR 33)ALRHP 34)ALRPH 35)ALHRP 36)ALHPR 37)AHLPR 38) AHLRP 39)AHRPL 40)AHRLP 41)AHPRL 42)AHPLR 43)APRLH 44)APRHL 45)APHLR 46)APHRL 47)APLRH 48)APLHR
49)LARPH 50)LARHP 51)LAPHR 52)LAPRH 53)LAHPR 54)LAHRP 55)LRPHA 56)LRPAH 57)LRHPA 58)LRHAP 59)LRAPH 60)LRAHP 61)LPRAH 62)LPRHA 63)LPAHR 64)LPARH 65)LPRAH 66)LPRHA 67)LHRAP 68)LHRPA 69)LHAPR 70)LHARP 71)LHPRA 72)LHPAR
73) PALRH 74)PALHR 75)PARLH 76)PARHL 77)PAHRL 78)PAHLR 79)PLRHA 80)PLRAH 81)PLHAR 82)PLHRA 83)PLARH 84)PLAHR 85)PHARL 86)PHALR 87)PHLRA 88)PHLAR 89)PHRLA 90)PHRAL 91)PRALH 92)PRAHL 93)PRLHA 94)PRLAH 95)PRHAL 96)PRHLA
97)HAPLR 98)HAPRL 99)HALPR 100)HALRP 101)HARLP 102)HARPL 103)HLPRA 104)HLPAR 105)HLARP 106)HLAPR 107)HLRAP 108)HLRPA 109)HPRAL 110)HPRLA 111)HPALR 112)HPARL 113)HPLAR 114)HPLRA 115)HRALP 116)HRAPL 117)HRPLA 118)HRPLA 119)HRLAP 120)HRLPA
Overall my prediction for being 120 arrangements for when n is 5 is correct. So now I will try to work out a general formula. I will do this by gathering all information to do with names with all different letters.
To represent 4*3*2*1 it can be expressed as 4 factorial. There is a button on most scientific calculators with have embedded this factorial button feature generally showing as an exclamation mark. All it does is save the time of having to put in to the calculator 1x2x3x4x5x6x7..... Ect. You just put in the number and press factorial and it will do 1x2x3... until it get to the number you put in. If I key in (lets say the number of letters all different) factorial 6 it gives me 720, with makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word and it continues to fall in that pattern.
Therefore a general formula for any word with all different letters can be written as
n!=a
WHY FORMULA WORKS
Using ‘Lucy’ as an example n=4
Using the formula n!=a
4!= 4*3*2*1=a
a=24
4 is in the formula because there are 4 letters and so there are 4 sets of arrangements for which each letter can begin with:
1)LUCY LUYC LCUY LCYU LYCU LYUC
2)ULCY ULYC UCLY UCYL UYLC UYCL
3)CLUY CLYU CULY CUYL CYLU CYUL
4)YLUC YLCU YCLY YCUL YULC YUCL
3 is in the formula because with in each set of arrangements there are 3 letters left which can follow:
1)LUCY LUYC 2)LCUY LCYU 3)LYUC LYCU
2 is in the formula because then there are only the last two letters which can be swapped
1)LUCY 2) LUYC
1 is in the formula because there is only one place to put the last letter
(1 arrangement)
1)LUCY
I will now investigate words that have repeated letters. Like in Emma it has 4 letters but 2 of letters are the same. I shall start of looking at words, which have 2 of the letters the same.
ARRANGEMENTS FOR YY
1)YY
There is one arrangement for this word, which is half the number of arrangements for when n=2 but all its letters different.
JO=2 arrangements YY=1 arrangement
ARRANGEMENTS FOR BBC
1)BBC 2)BCB 3)CBB
There are 3 different arrangements for this word, which is half the number of arrangements for when n=3 with all different letters.
BBC=3 arrangements JIM=6 arrangements
Having already investigated the number of arrangements for the word Emma, I have compared the number of arrangements with the name Lucy. I have found that Emma has half the number of arrangements than LUCY, which has all different letters.
EMMA=12 Arrangements LUCY=24 Arrangements
Prediction
Now I shall investigate the number of arrangements for when n=5 with 2 of the letters the same, I shall use the name DAVID.
Judging from the above results I can see that the number of arrangements for a word with 2 letters the same, will have half the number of arrangements than the word with the same total number of letters but with all different letters. Therefore as the name RALPH has 120 arrangements so as DAVID has 2 letters the same a shall be half of RALPH. Therefore as Ralph has 120 arrangements the name DAVID will have 60 arrangements. 120/2=60
ARRANGEMENTS FOR DAVID
1)DAVID 2)DAVDI 3)DADVI 4)DADIV 5)DAIVD 6)DAIDV 7)DDVIA 8)DDVAI 9)DDAIV 10)DDAVI 11)DDIVA 12)DDIAV 13)DVDIA 14)DVDAI 15)DVADI 16)DVAID 17)DVIDA 18)DVIAD 19)DIDVA 20)DIDAV 21)DIVAD 22)DIVDA 23)DIAVD 24)DIADV
25)AVDID 26)AVDDI 27)AVIDD 28)ADDVI 29)ADDIV 30)ADIVD 31)ADIDV 32)ADVID 33)ADVDI 34)AIDDV 35)AIDVD 36)AIVDD
37)VDDAI 38)VDDIA 39)VDIAD 40)VDIDA 41)VDADI 42)VDAID 43)VADDI 44)VADID 45)VAIDD 46)VIDDA 47)VIDAD 48)VIADD
49)IDDVA 50)IDDAV 51)IDVAD 52)IDVDA 53)IDAVD 54)IDADV 55) IVDDA 56)IVDAD 57)IVADD 58)IADVD 59)IADDV 60)IAVDD
As you shown, my prediction for being 60 arrangements for the 5-lettered name is right. Now I shall compare the words with 2 letters the same and words with all different letters.
OBSERVATIONS
The main comparison I noticed between a word with all different letters and a word with 2 of its letters the same, is that a for a word with all different letters is double the amount of arrangements for a word with the same amount of letters but 2 of its letters the same. So a general formula for a word with 2 letters the same is
n!/2=a
n= number of letters in word.
a= number arrangements.
Now I shall look at words, which have 3 of its letters the same.
ARRANGEMENTS FOR XXX
1) XXX
There is only 1 arrangement for this 3-lettered word, which has 3 letters the same. This is a third of the number of arrangements than a word with the same amount of letters but only 2 of its letters the same, and 1 sixth of the number of arrangements for a word the same number of letters but with all different letters.
ARRANGEMENTS FOR XXXY
1) XXXY 2) XXYX 3) XYXX 4) YXXX
There are 4 arrangements for a 4-lettered word, which has 3 letters the same. This is a third of the number of arrangements than a word with the same amount of letters but only 2 of its letters are the same, and one sixth the number of arrangements for a word with the same amount of letters but all different letters.
Prediction for XXXYZ
From some of the above results I could probably predict that the number of arrangements for this word, would be one sixth the number of arrangements for a word with the same number of letters but having all its letters different.
ARRANGEMENTS FOR XXXYZ
1) XXXYZ 2) XXXZY 3) XXYZX 4) XXZYX 5) XYZXX 6) XZYXX 7) ZXXXY 8) ZXXYX 9) ZXYXX 10) ZYXXX 11) YXXXZ 12) YXXZX 13) YXZXX 14) YZXXX 15) XZXYX 16) XZXXY 17) XYXZX 18) XXZXY 19)
My prediction was right I shall now compare all my results for words having 3 letters the same with words having 2 of its letters the same and words with all its letters different.
Observations
The main comparison between a word with 3 letters the same and word with 2 of its letters the same and a word with all different letters, is that the words with 3 letters the same has one third the amount of arrangements than a word with the same amount of letters but 2 of its letters the same. Also the words with 3 letters the same has one sixth the amount of arrangements than words with the same amount of letters but different all different letters. Therefore I show a formula for a word with 3 of its letters the same.
n!/6
Now I shall investigate on words, which have 4 of its letters the same.
Number of arrangements for AAAA
1) AAAA
There is only one arrangement for this word, which has all four of its letters the same. This one twentieth of the number of arrangements than a word with the same amount of letters but with all different letters. It is one twelfth of the number of arrangements for the word with the same amount of letters but 2 letters the same. Finally it is a quarter of the number of arrangements for a word with the same number of letters but with 3 of its letters the same.
NUMBER OF ARRANGEMENTS FOR MMMMA
1)MMMMA 2)MMMAM 3)MMAMM 4)MAMMM 5)AMMMM
There are 5 arrangements for this word, which has 4 letters the same, this is one twentieth the amount of arrangements than a word with the same amount of letters but all the letters different. It is one twelfth of the number of arrangements for the word with the same amount of letters but 2 of its letters the same. Finally it is a quarter of the number of arrangements for a word with the same number of letters but with 3 of its letters the same.
I will now gather the information that I have to do with words with 4 of the same letters and compare them to words that have all different letters, words with 2 of its letters the same, and words that have 3 letters of its letters the same.
Observations
The main comparison between a word with 4 letters the same, word with 3 of its letters the same, a word with 2 of its letters the same and a word with all different letters which has the same total of letters, is that the words with 4 letters the same has one twenty fourth the amount of arrangements than a word with all different letters, it has 1 twelfth the amount of arrangements for a word with the same amount of letters but 2 of its letters the same. Also it is a quarter the amount of arrangements than a word with the same amount of letters but 3 of its letters the same. Therefore I can show a formula for a word with 3 of its letters the same. I know this because the formula for a word with all different letters is n! and as the number of arrnagments for a word with all different letters has 24 times the amount of arrangements for a word with the same number of letters but 4 of its letters the same, so the formula must be for a word with 4 letters the same
n!/24
