Also we can use this method to find the different arrangements of LUCY ‘s name.
3) Investigate the number of the different arrangements of the letter s of LUCY ‘s name.
1) EHSAN
EAHNS EHANS ENAHS ESAHN
EAHSN EHASN ENASH ESANH
EANHS EHNAS ENHAS ESHNA 6x4=24
EANSH EHNSA ENHSA ESHAN There is 24 different
EASHN EHSAN ENSAH ESNAH arrangements for EASNH EHSNA ENSHA ESNHA letter E.
HAENS HEANS HNAES HSAEN
HAESN HEASN HNASE HSANE 6x4=24
HANES HENAS HNEAS HSEAN There is 24 different
HANSE HENSA HNESA HSENA arrangements for
HASEN HESAN HNSAE HSNAE letter H.
HASNE HESNA HNSEA HSNEA
SAEHN SEAHN SHAEN SAEHN
SAENH SEANH SHAEN SAENH 6x4=24
SANEH SEHAN SHEAN SAHEN There is 24 different
SANHE SEHNA SHENA SAHNE arrangements for
SAHEN SENAH SHNAE SANEH letter S.
SAHNE SENHA SHNEA SANHE
AEHNS AHENS ANEHS ASEHN
AEHSN AHESN ANESH ASENH 6x4=24
AENHS AHNES ANHES ASHEN There is 24 different
AENSH AHNSE ANHSE ASHNE arrangements for
AESHN AHSEN ANESH ASNEH letter A.
AESNH AHSNE ANEHS ASNHE
NAEHS NEAHS NHAES NSAEH 6x4=24
NAESH NEASH NHASE NSAHE
NAHES NEHAS NHEAS NSEAH There is 24 different NAHSE NEHSA NHESA NSEHA arrangements for
NASEH NESAH NHSAE NSHAE letter N.
NASHE NESHA NHSEA NSHEA
When we add all of them up we get 120, which is the total combination of the letter EHSAN.
2) JACK
JACK ACJK CAJK KACJ
JAKC ACKJ CAKJ KAJC
JCAK AKCJ CJKA KCAJ
JCKA AKJC CJAK KCJA
JKAC AJCK CKAJ KJAC
JKCA AJKC CKJA KJCA
Each letter has 6 different arrangements and we have 4 different letters, in order to find out how many different Arrangements we have to multiply 4
by 6,which is 24.
3) ALI
AIL IAL LAI
ALI ILA LIA
Each letter has 2 different arrangements and we have 3 different letters and in order to find out different Arrangements we have to multiply 2 by 3,which is 6.
4) ET
ET
TE
Each letter has one different arrangement and we have 2 different letters and in order to find out different arrangements we have to multiply 1 by two, which, is 2.
But as we can see, this method takes lots of time and it is not accurate either so I tried to find another formula, which is a lot easier and quicker.
First in order to find the formula, I want to see in how many ways I can arrange letters chosen from the 3 letters E, A and N.
E A N
E or A or N A or N N
3 X 2 X 1 = 6
1-E A N
2-E N A
3-A E N
4-A N E
5-N A E
6-N E A
The first letter can be chose in 3 ways and the second letter can be chosen in 2 ways and the last place in (1) way. And so I can calculate the number of arrangement of my 3 letters word which is obviously 3x2x1= 6.
So from top experiment I can prove the bottom theory.
Suppose we wish to arrange N objects the first abject can be chosen in (n) ways the second object can be chosen in (n - 1) ways, the third object can be chosen in (n - 2) ways… and so on until the last object which can be written in (1) ways. So the formula to find the different arrangements of a letter is:
n (n - 1) (n – 2) (n - 3) (n – 4) …. (n - r + 1)
OR
n!
ABCDEFGHIJ: 10 X 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1 = 3628800
ABCDEFGHI: 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1 = 362880
ABCDEFGH: 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1 = 40320
ABCDEFG: 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040
ABCDEF: 6 x 5 x 4 x 3 x 2 x 1 = 720
ABCDE: 5 x 4 x 3 x 2 x 1 = 120
ABCD: 4 x 3 x 2 x 1 = 24
ABC: 3 x 2 x 1 = 6
AB: 2 x 1= 2
A: 1
0! : 1 I will prove this at the end.
However we can only use this formula to find the different arrangements of unlike things and we cannot use it to find the different arrangements of like things so I want to now find a formula, which can find arrangements of both like and unlike things.
Suppose the n unlike letters a1 a2 a3 … b1 b2 … c1 c2 … are arranged in a row, then the number of possible arrangements is n!
If we consider these n ! arrangements and remove suffixes from the letters ‘a’ we shall have written down the same arrangement several times for example for arrangements
a1 … a3 … a2
a2 … a1 … a3
a3 … a1 … a2
will all be the same once the suffixes are removed and certainly, we shall have repeated this particular arrangement r ! where r is the number of a ’s. In the same way we can consider the b ‘s and c ‘s.
For this reason the number of arrangements of n things, p of one kind, q of another, r of another … is
n ! or (p + q + r)!
p ! q ! r !
I want to see in how many ways I can arrange two different letters chosen from the five letters E, H, S, and N.
EH The first letter can be chose in 5 ways
HE and the second letter can be chosen in
ES 4 ways.
SE Suppose we wish to arrange r objects
EA chosen from n unlike object:
AE The first object can be chosen in n
EN ways the second object can be chosen
NE in (n - 1) ways, the third object can be
HS chosen in (n - 2) ways… and so on until
SH the 4th object which can be written in
HA 20 (n – r + 1) ways.
AH So the formula to find permutations
HN of r objects chosen from n unlike objects
NH is:
SA n (n - 1) (n – 2) (n - 3) (n – 4) …. (n - r + 1)
AS
SN n (n - 1) (n – 2) … (n - r + 1) n !
NS = =
AN (n – r)(n – r – 1) … 2 x 1 (n - r) !
NA
We usually say that the number of permutations of r objects selected from
n !
n unlike objects is “P r where P r =
(n - r) !
Thus in the case of arranging two letters chosen from the five letter E,H,S,
5!
A and mentioned above, this gives 5P2 =
3!
= 20 as requirements
we know that there are n! arrangements of n objects
chosen from n objects( is n! ). Thus “P,, must equal n !
n!
i.e. = n!
(n - n) !
and we therefore define 0! as 1.
4) XX …XXYY...YY
Investigate the number of different arrangements of the letters.
We assume we have n number of X and m number of Y
XX…XXX YY…YYY
n m
According to my pervious investigation I can write the number of different arrangements of X ‘s and Y ‘ as below.
(n + m)!
n! m!
