Part 2
Introduction
In this investigation, I am going to find out the different arrangements of the letters of the name EMMA
Here are the possibilities for the name EMMA
EMMA
EMAM
EAMM
MEMA
MEAM
MMEA
MMAE
MAEM
MAME
AMME
AMEM
AEMM
First of all I took the letter M from the name EMMA then wrote the second letter being E and after the two other possibilities being M and A. Then after I used the same letter M, wrote the second letter down being M and wrote the two possibilities again being E and A. I took another letter which I used that to be the second letter, keeping the first letter the same in till I have found the rest of the possibilities in the letter. Wrote down the 2 other possibilities. There were six possibilities for the word beginning with M. I repeated the process with the letters A and M. I found out that there were 6 possibilities for the letter M and 3 possibilities for the letters A and E. The reason for the letters A and E have only 3 possibilities is because there were two M’s in the name EMMA, so I could not of used M twice in the same word. Example: If I hade the name EMMA I would not have been allowed to use the name again because of the two M’s. I would not be allowed to swoop them over because they would not make any difference and same goes with the letter A, I wont be allowed to swoop the two M’s over. It wont be fair test.
I am now going to try to find a pattern in the possibilities of combinations of different letters.
a) Firstly with one letter.
A
There is one permutation with one letter.
b) With two letters.
AB BA
There are two possibilities with two different letters.
c) With three letters.
ABC BAC CBA
ACB BCA CAB
There are six permutations with three different letters.
d) With four letters.
ABCD BACD CABD DABC
ABDC BADC CADB DACB
ACBD BCAD CBAC DBAC
ACDB BCDA CBCA DBCA
ADCB BDAC CDAB DCAB
ADBC BDCA CDBA DCBA
There are twenty-four permutations with four different letters.
1! = 1 x 1 =1
2! = 1 x 2 =2
3! = 1 x 2 x 3 =6
4! =1 x 2 x 3 x 4 =24
From this table I predict that with five letters there will be 120 permutations.
5! = 1 x 2 x 3 x 4 x 5 = 120
e) With five letters.
ABCDE BACDE CABDE DABCE EABCD
ABCED BACED CABED DABEC EABDC
ABDCE BADCE CADBE DACBE EACBD
ABDEC BADEC CADEB DACEB EACDB
ABECD BAECD CAEBD DAEBC EADBC
ABEDC BAEDC CAEDB DAECB EADCB
ACBDE BCADE CBADE DBACE EBACD
ACBED BCAED CBAED DBAEC EBADC
ACDBE BCDAE CBDAE DBCAE EBCAD
ACDEB BCDEA CBDEA DBCEA EBCDA
ACEBD BCEAD CBEAD DBEAC EBDAC
ACEDB BCEDA CBEDA DBECA EBDCA
ADBCE BDACE CDABE DCABE ECABD
ADBEC BDAEC CDAEB DCAEB ECADB
ADCBE BDCAE CDBAE DCBAE ECBAD
ADCEB BDCEA CDBEA DCBEA ECBDA
ADEBC BDEAC CDEAB DCEAB ECDAB
ADECB BDECA CDEBA DCEBA ECDBA
AEBCD BEACD CEABD DEABC EDABC
AEBDC BEADC CEADB DEACB EDACB
AECBD BECAD CEBAD DEBAC EDBAC
AECDB BECDA CEBDA DEBCA EDBCA
AEDBC BEDAC CEDAB DECAB EDCAB
AEDCB BEDCA CEDBA DECBA EDCBA
