2 Arrangements And A 3 Letter Word example essay topic

Emm as Dilemma In my investigation I am going to investigate the number of different arrangements of letters in a word. e.g. Tim Is one arrangement Mit Is another First I am going to investigate how many different arrangements in the name LUCY, which has no letters the same. LUCY LUC LYCUS LUC LC YU LC UY ULC Y UGLY ULC ULY C UCYLThere are 4 different letters and there are 24 different arrangements. SAM SMA MSA MAS ASM AMS There are 3 different letters in this name and 6 different arrangements. JO OJ There are 2 different letters in this name and there are 2 different arrangements. UCL Y CUL Y UCL YUL C YL CU Y LUC CYL U C YUL CURL C ULY CLUB Table of Results Number of Letters Number of Different Arrangements 2 2 3 6 4 24 5 120 6 720 7 5040 From the table of results I have found out that a 2 letter word has 2 arrangements, and a 3 letter word has 6. Taking for example a 3 letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements.

In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4! , which is called 4 factorial which is the same as 4 x 3 x 2. So, by using factorial (!) I can predict that there will be 40320 different arrangements for an 8 letter word. The formula for this is: n! = a Where n = the number of letters in the word and a = the number of different arrangements. Now I am going to investigate the number of different arrangements in a word with 2 letters repeated, 3 letters repeated and 4 letters repeated.

EMMA A MME A MEM EXAM A EMM E AMM MME A MMA E MEM A MAME MEAN MEM 4 letter word, 2 letters repeated, 12 different arrangements. MUM MMU UMM 3 letter word, 2 letters repeated, 3 different arrangements. PQ PMQMM PM MQM PMM MQ QP QM PMM QM MPM QM MMP MPQMM MP MQM MPM MQ MQ PMM MQM PM MQM MP MMP QM MMQMP PQ QP MMP Q QP 5 letter word, 3 letters repeated, 20 different arrangements. S MSM M MMS M S 4 letter word, 3 letters repeated, 4 different arrangements. K K 5 letter word, 4 letters repeated, 5 different arrangements.

Table of Results No of letters (n) No letters repeated Same letter repeated 2 x (p) Same letter repeated 3 x (p) Same letter repeated 4 x (p) Same letter repeated 5 x (p) 3 6 3 1 0 0 4 24 12 4 1 0 5 120 60 20 5 1 6 720 360 120 30 6 7 5040 2520 840 210 42 I have worked out that if you do say 5! = 120, to find out how many different arrangement in a 3 letter word it would be 5! Divided by 6 = 20, so, a 6 letter word with 4 letters repeated would be 6! Divided by 24 = 30, as you can see in the No letters repeated column these are the numbers we are dividing by: 2 letters the same = n!

(21 = 2) 3 letters the same = n! (32 x 1 = 6) 4 letters the same = n! (43 x 21 = 24) 5 letters the same = n! (54 x 32 x 1 = 120) 6 letters the same = n!

(65 x 43 x 21 = 720) From this I have worked out the formula to fine out the number of different arrangements: n! = the number of letters in the word p! = the number of letters the same Now I am going to investigate the number of different arrangement for words with 2 or more letters the same like, aabb, bb, or. This is a 4 letter word with 2 letters the same, there are 6 different arrangements: xxyyI am going to use the letters x and y (any letter) yy xy xy yx xy xy yx yx yx yyxxThis is a 5 letter word yy yy xxyxy xyxyx xyxxy xyyxx yy y y yxyxx yxxyxThere are 10 different arrangements In the above example there are 3 x's and 2 y's As each letter has its own number of arrangements i.e. there were 6 beginning with x, and 4 beginning with y, I think that factorial has to be used again. As before, the original formula: n! = the number of letters in the word p! = the number of letters the same From this I have come up with a new formula The number of total letters factorial, divided by the number of x's, y's etc factor ised and multiplied. For the above example: A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's) So: 12 x 34 (12) x (12) = 24 4 = 6 different arrangementsA five letter word like bb; this has 4 a's and 1 b (4 x's and 1 y) So: 12 x 34 x 5 (12 x 34) x (1) = 120 24 = 5 different arrangements A five letter words like abcde; this has 1 of each letter (no letters the same) So: 12 x 34 (11 x 11 x 11) = 24 1 = 24 different arrangementsA five letter word like bb; this has 3 a's and 2 b's (3 x's and 2 y's). So: 12 x 34 x 5 (12 x 3) x (12) = 120 12 = 10 different arrangements This shows that my formula works: n! = the number of letters in the word x! y! = the number of repeated letters the same By Katherine Bond 11 W.