U L CYUL
Y L U CYLU
L C YULC The name Lucy twenty-four
U C L YUCL different arrangements.
Y U L YCUL
C L U YCLU
U C YLUC
L C U YLCU
Emma
M A EMMA
M A M EMAM
E A M M EAMM
M A MEMA M has more arrangements than E
E A M MEAM or A because the name Emma has
M E M MAEM two M’s in.
A M E MAME
E A MMEA
M A E MMAE
E M AMEM The name Emma has twelve
M M E AMME different arrangements.
A E M M AEMM
From these results I can construct a table.
The number of arrangements = (number of arrangements for the name with 1 less letter) multiplied by number of letters in current name.
Using this formula I can predict the number of arrangements for names with more letters.
I will now use the branching out method to see whether my prediction for the name Mandy is right.
I predict Mandy will have120 different arrangements.
D Y MANDY
N Y D MANYD
A N Y MADNY
D Y N MADYN
D N MAYDN
Y N D MAYND
A D MNYAD
Y D A MNYDA
A Y MNDAY
D Y A MNDYA
D Y MNADY
N A Y D MNAYD
M A N MDYAN
Y N A MDYNA
A Y MDNAY
D N Y A MDNYA
N Y MDANY
A Y N MDAYN
A N MYDAN
D N A MYDNA
D A MYNDA
Y N A D MYNAD
D N MYADN
A N D MYAND
D Y ANMDY
M Y D ANMYD
N M Y ANDMY
D Y M ANDYM
D M ANYDM
Y M D ANYMD
N M ADYNM
Y M N ADYMN
M Y ADNMY
N Y M ADNYM
N Y ADMNY
D M Y N ADMYN
A M N AYDMN
D N M AYDNM
A M AYNAM
Y N M A AYNMA
N D AYMND
M D N AYMDN
Y N AMDYN
D N Y AMDNY
D Y AMNDY
M N Y D AMNYD
D N AMYDN
Y N D AMYND
D Y NAMDY
M Y D NAMYD
A M Y NADMY
D Y M NADYM
D M NAYDM
Y M D NAYMD
A M NDYAM
Y M A NDYMA
M Y NDAMY
A Y M NDAYM
A Y NDMAY
D M Y A NDMYA
N M A NYDMA
D A M NYDAM
D M NYADM
Y A M D NYAMD
A D NYMAD
M D A NYMDA
Y A NMDYA
D A Y NMDAY
D Y NMADY
M A Y D NMAYD
D A NMYDA
Y A D NMYAD
N Y DAMNY
M Y N DAMYN
A M Y DANMY
N Y M DANYM
N M DAYNM
Y M N DAYMN
A M DNYAM
Y M A DNYMA
M Y DNAMY
A Y M DNAYM
A Y DNMAY
N M Y A DNMYA
D M A DYNMA
N A M DYNAM
N M DYANM
Y A M N DYAMN
A N DYMAN
M N A DYMNA
Y A DMNYA
N A Y DMNAY
N Y DMANY
M A Y N DMAYN
N A DMYNA
Y A N DMYAN
N D YAMND
M D N YAMDN
A M D YANMD
N D M YANDM
N M YADNM
D M N YADMN
A M YNDAM
D M A YNDMA
M D YNAMD
A D M YNADM
A D YNMAD
N M D A YNMDA
Y M A YDNMA
N A M YDNAM
N M YDANM This proves that my
D A M N YDAMN prediction that the name
A N YDMAN Mandy can be arranged
M N A YDMNA 120 different ways.
D A YDNYA
N A D YDNAY
N D YDANY
M A D N YDAYN
N A YMDNA
D A N YMDAN
I will now investigate word with two or more letters the same.
AA
AA
Can only be arranged as AA
ANN
ANN NAN NNA
ANN can be arranged three different ways
EMMA
EMMA AEMM MAEM MMAE These are the 12
EMAM AMEM MEAM MAME different arrangement of
EAMM AMME MMEA MEMA letters in the name Emma.
I will now show my results for names that have two letters the same in a table.
Using my table I can predict the number of different arrangements for a name that has two letters the same.
I predict the name Danny will have 60 different arrangements
DANNY
DANNY YDANN NYDAN NNYDA ANNYD
DANYN YDNAN NYDNA NNYAD ANNDY
DAYNN YDNNA NYAND NNADY ANDYN
DYNNA YANND NYADN NNAYD ANDNY
DYNAN YANDN NYNDA NNDYA ANYDN
DYANN YADNN NYNAD NNDAY ANYND
DNANY YNAND NANYD AYNND
DNAYN YNADN NAYDN AYNDN
DNYAN YNDAN NADNY AYDNN
DNNYA YNNDA NANDY ADYNN
DNNAY YNNAD NAYND ADNYN
DNYNA YDANY NADYN ADNNY
NDYNA
NDAYN
NDNYA There is sixty different
arrangements of the letters
NDYAN in the name Danny. My prediction
was right
NDANY
NDNAY
I am now going to investigate the different arrangements of different numbers of X’s and Y’s in a word.
xy
xy yx There is only two ways to arrange xy
Xyy
Xyy yxy yyx There is three ways to arrange xyy
Xyyy
Xyyy yxyy yyxy yyyx There is four ways xyyy
Can be rearranged.
From these results I have constructed a table.
I will now investigate more word that only have X’s and Y’s.
Xxyy
Xxxyy Xxxxy
Xxyy
Xxyy xyxy yxxy
Xyyx yxyx yyxx
There are six arrangements for XXYY
Xxxyy
Xxxyy xxyxy xxyyx xyxyx xyxxy
Xyyxx yyxxx yxxxy yxyxx yxxyx
There are ten different arrangements for this sequence
Xxxxy
Xxxxy xxxyx xxyxx xyxxx yxxxx
In xxxyy there is 3 X’s and 2 Y’s, which makes 5 unknown. Each letter has so many arrangements that begin with that letter so I have to factorial.
E.g. A five-letter word like xxxyy; this has 3 X’s and 2 Y’s.
So: 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10
(no of letters) (no of X’s)(no of Y’s)
A four letter word like xxyy; this has 2 X’s and 2 Y’s
So: 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6
Five letter words like vwxyz; this has 1 of each letter (no letters the same)
So: 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24
A= ni/6
Formula
The formula is easier to show using numbers
Formula for all different number:
a=ni
formula when 2 number are the same
a=ni/2
formula when 3 numbers are the same
a=ni/6
Let’s put them is this way:
n represent the number of figures of a number
x represents the divided number in the formula
X is equal the last x*n, so I predict that the formula for 4 same number of a number is:
A= ni/24
4 fig, one arrangement.
a=n/24=(1*2*3*4)/24=1so the formula works
(4)
if a figure has two pairs of three same numbers the formula is (1*2*3*4*5*6*7*8)/1*2*3*4*1*2*3*4=70
Use this formula, we can find out the total arrangements of all numbers and letters.
3 letters the same = n! ( 3x2x1 = 6)
4 letters the same = n! (4x3x2x1 = 24)
5 letters the same = n! (5x4x3x2x1 = 120)
6 letters the same = n! (6x5x4x3x2x1 = 720)
From this I have worked out the formula to find out the number of different arrangements:
n! = the number of letters in the word
p! = the number of letters the same
a=n!/p!(formula)
In conclusion: I have learnt how to systematically rearrange different numbers and letters in a sequence.
I have also developed my factorial skills.
