- kenir different ways starting with the letter ‘i’.
I have looked at my results and I have found a pattern:
1 = 1
2 = 2 x 1
6 = 3 x 2 x 1
24 = 4 x 3 x 2 x 1
120 = 5 x 4 x 3 x 2 x 1
So I predict that a six-letter word would have 720 different arrangements.
720 = 6 x 5 x 4 x 3 x 2 x 1
SELINA – selina - slinae - sinael - snaeli - saelin
- selian - slinea - sinale - snaeil - saelni
- selnia - sliane - sinela - snalei - saeinl
- selnai - sliaen - sineal - snalie - saeiln
- selani - sliean - sinlea - snaile - saenil
- selain - sliena - sinlae - snaiel - saenli
--------- --------- --------- --------- ---------
- seinal - slnaei - siaenl - snelia - salien
- seinla - slnaie - siaeln - snelai - saline
- seialn - slneai - sialen - sneila - salein
- seianl - slneia - sialne - sneial - saleni
- seilan - slniea - sianle - sneali - salnei
- seilna - slniae - sianel - sneail - salnie
--------- --------- --------- --------- ---------
- senail - slaein - sielna - snliae - sainel
- senali - slaeni - sielan - snliea - sainle
- senlai - slaien - sienal - snlaie - saieln
- senlia - slaine - sienla - snlaei - saienl
- senile - slanie - siealn - snleai - sailne
- senial - slanei - sieanl - snleia - sailen
--------- --------- --------- --------- ---------
- sealin - sleina - silnae - sniael - saneil
- sealni - sleian - silnea - sniale - saneli
- seailn - slenai - silane - snieal - sanlie
- seainl - slenia - silaen - sniela - sanlei
- seanli - sleani - silean - snilea - sanile
- seanil - sleain - silena - snilae - saniel
There are 120 different ways of arranging the word ‘selina’ beginning with the letter ‘s’. There are 6 letters in the word, so 6 x 120 = 720. Which is how many ways I predicted it would have.
A formula can be made to predict the number of different combinations for any number of letters in a word. If the number of letters is n then the formula will be n(n-1)(n-2)…
This works out because there are n possible letters for the first slot, then only n-1 left for the second slot, and so on, down to the last letter, where you only have one choice.
The formula is;
n(n-1)(n-2)(n-3)(n-4)…
So in a 4-letter word n = 4. So the equation will be 4 x 3 x 2 x 1, which is what I have been doing, but now I have a formula for any number of letters in a word.
I started to play with the calculator and I noticed a key ‘n!’ I played around and worked out that if you type in the number of letters of the word and then press the ‘n!’ key you get the answer of how many arrangements you get. I checked this by doing 5! = 120, which is correct. Therefore ‘n!’ on the calculator is the same as my formula.
Then I looked at words with repeats in them. But at first I am only going to have one letter repeat twice in each word.
BOB - bob
- bbo
- obb
EMMA - emma - amem - mame
- emam - aemm - maem
- eamm - mema - mmea
- amme - meam - mmae
CLARA - clara - lcraa - aacrl - alcar
- claar - lcara - aalrc - alarc
- clraa - lcaar - aalcr - alacr
--------- --------- - aarlc - alrac
- calra - lrcaa - aarcl - alrca
- calar - lraca ---------
- caalr - lraac - racla
- caarl --------- - racal
- carla - aracl - raalc
- caral - aralc - raacl
--------- - arlca - ralac
- craal - arlac - ralca
- crala - arcla ---------
- crlaa - arcal - rclaa
--------- --------- - rcala
- larac - acrla - rcaal
- larca - acral ---------
- lacar - acarl - rlaca
- lacra - acalr - rlaac
- laarc - aclar - rlcaa
- laacr - aclra ---------
--------- - aaclr - alcra
In looking at both results I noticed that a 3-letter word with no repeats in it has 6 different arrangements and a 3-letter word with 1 letter repeated twice has 3 different arrangements.
6/2 = 3
A 4-letter word with no repeats has 24 different arrangements and a 4-letter word with 1 letter repeated twice has 12 different arrangements.
24/2 = 12
I have worked out that a word with 1 letter repeated twice has half the number of different arrangements as a number with no repeats.
Then I tried words with one letter repeated 3 times.
AAA - aaa
ABBB - abbb
- babb
- bbab
- bbba
ABCCC - abccc - cbcac
- acbcc - cbcca
- accbc - cabcc
- acccb - cacbc
--------- - caccb
- baccc - ccbac
- bcacc - cccba
- bccac - ccbca
- bccca - cccab
--------- - ccabc
- cbacc - ccacb
Looking at these results and my last results I have noticed that using my first formula then dividing it by the number of repeated letters you could work out any word with repeated letters.
n(n-1)(n-2)(n-3)… n=total number of letters in the word
x(x-1)(x-2) x=number of repeated letters
So for a 6-letter word with one letter repeated 3 times the formula is;
6(6-1)(6-2)(6-3)(6-4)(6-5)
3(3-1)(3-2) = 120 different arrangements
If a calculator is used the formula is;
n!
x!
Which for this calculation is;
6!
3! = 120 different arrangements
So I predict that a 6-letter word with one letter repeated 3 times will have 120 different arrangements.
ABCDDD - abcddd - bcaddd - cdabdd - dbacdd - ddabcd
- abdcdd - bcdadd - cdbadd - dbadcd - ddabdc
- abddcd - bcddad - cdbdda - dbaddc - ddadbc
- abdddc - bcddda - cdbdad - dbcadd - ddadcb
- acbddd - bdacdd - cdadbd - dbcdad - ddacdb
- acdbdd - bdcadd - cdaddb - dbcdda - ddacbd
- acddbd - bdcdda - cddbad - dbdacd - ddbacd
- acdddb - bdcdad - cddabd - dbdcad - ddbadc
- adbcdd - bdadcd - cddadb - dbddca - ddbcad
- adcbdd - bdaddc - cddbda - abddac - ddbcda
- adcdbd - bddcad - cdddab - dbdadc - ddbdac
- adcddb - bddacd - cdddba - dbdcda - dddbca
- adbdcd - bddadc - dabcdd - dcabdd - ddcabd
- adbddc - bddada - dabdcd - dcadbd - ddcadb
- addcbd - bdddac - dabddc - dcaddb - ddcbad
- addbcd - bdddca - dacbdd - dcbadd - ddcbda
- addbdc - cabddd - dacdbd - dcbdad - ddcdab
- addcdb - cadbdd - dacddb - dcbdda - ddcdba
- adddbc - caddbd - dadbcd - dcdabd - dddabc
- adddcb - cadddb - dadcbd - dcdbad - dddacb
- bacddd - cbaddd - daddcb - dcddba - dddbac
- badcdd - cbdadd - daddbc - dcddab - dddbca
- baddcd - cbddad - dadbdc - dcdadb - dddcab
- badddc - cbddda - dadcdb - dcdbda - dddcba
A 6-letter word with one letter repeated 3 times has 120 different arrangements. So the formula is correct.
Then I checked that it would work if in a word there were 2 letters repeated twice.
AABB - aabb
- abab
- abba
- baab
- baba
- bbaa
ABBCC - abbcc - bbacc - bacbc - bccab - cbbca
- abcbc - bbcac - baccb - bccba - cbcab
- abccb - bbcca - bcabc - cacbb - cbcba
- accbb - cabbc - bcacb - cbabc - ccabb
- acbcb - cabcb - bcbac - cbacb - ccbab
- acbbc - babcc - bcbca - cbbac - ccbba
4! = 24
24
6 = 4
The factors of 4 are 2 and 2
Which are both letters repeated twice,
n(n-1)(n-2)(n-3)…
(x1(x-1)(x-2)…*x2(x-1)(x-2)…)
So for a 5-letter word with 2 letters repeated twice the equation will be,
5(5-1)(5-2)(5-3)(5-4)
(2(2-1)*2(2-1))
Or on a calculator,
5!
(2!*2!) = 30 different arrangements
With all this information the total number of combinations can now be calculated for any word with any number of letters with any number of repeated letters too. All these formulae have a name, factorial and a notation, n!.
