I will now go through the different arrangements for names or word arrangements that do not have any repeats and are one, two, three, four and five letters long:
A
JO OJ
MAX MXA AMX
AXM XMA XAM
MIKE MIEK MKIE MKEI MEIK MEKI
IMKE IMEK IKME IKEM IEMK IEKM
KMIE KMEI KIME KIEM KEMI KEIM
EMIK EMKI EIMK EIKM EKMI EKIM
MINTY MNITY MTINY MYINT
MINYT MNIYT MTIYN MYITN
MITNY MNTIY MTNIY MYNIT
MITYN MNTYI MTNYI MYNTI
MIYNT MNYIT MTYIN MYTIN
MIYTN MNYTI MTYNI MYTNI
IMNTY INMTY ITMNY IYMNT
IMNYT INMYT ITMYN IYMTN
IMTNY INTMY ITNMY IYNMT
IMTYN INTYM ITNYM IYNTM
IMYNT INYMT ITYMN IYTMN
IMYTN INYTM ITYNM IYTNM
NMITY NIMTY NTMIY NYMNT
NMIYT NIMYT NTMYI NYMTN
NMTIY NITMY NTIMY NYIMT
NMTYI NITYM NTIYM NYITM
NMYIT NIYMT NTYMN NYTMN
NMYTI NIYTM NTYNM NYTNM
TMINY TIMNY TNMIY TYMIN
TMIYN TIMYN TNMYI TYMNI
TMNIY TINMY TNIMY TYIMN
TMNYI TINYM TNIYM TYINM
TMYIN TIYMN TNYMI TYNMI
TMYNI TIYNM TNYIM TYNIM
YMINT YIMNT YNMIT YTMIN
YMITN YIMTN YNMTI YTMNI
YMNIT YINMT YNIMT YTIMN
YMNTI YINTM YNITM YTINM
YMTIN YITMN YNTMI YTNMI
YMTNI YITNM YNTIM YTNIM
I displayed each number of letters in the way of a tree diagram, each time changing the last letter of the word, then working my way back along the word, until I eventually had the letters in reverse order. I then recorded the results in this table:
Here is how I found the rule of working out the number of arrangements for a word with no repeated letters. If you multiply the number of letters of each word followed by the next in the series, you will get the number of different arrangements for the next number in the series you are looking for. So, If you were looking for the number off different arrangements for the 7th term in the series, you would multiply 1x2x3x4x5x6x7 (number of letters) which would equal 5040 (number of different arrangements). I then discovered you could use the “factorise” (!) button on the calculator, so you could simply type in 7! And reach the same answer. The correct formula using algebra would therefore be n!, as the factorise button multiplies the number you are trying to find by all the numbers before it.
After finding the formula for this question I moved onto names with 1 set of repeated words such as Emma. Again I used the same method as I did to find the previous number arrangements for Lucy, using one, two and three letters, with one letter being repeated each time:
AA
AAB ABA BAA
AABC AACB ABAC
ABCA ACAB ACBA
BACA BAAC BCAA
CAAB CABA CBAA
I used the same method that I used before and discovered that words with one set of repeated letters were exactly the same as the previous results, except that the number of arrangements was halved when the letters were repeated. Therefore the algebraic formula would be n!/2. I then had to find a rule that could be applied to words with no repeated letters, two repeated letters and then three repeated letters. According to my previous formulas, an arrangement of words with three repeated letters should have an algebraic formula of n!/3. I decided to display my results in the following table:
Whilst looking for different rules I found that you can also get the same answer by using the algebraic expression n!/r!, which means that the number of letters factorized divided by the number of repeated letters.