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Emmas Dilemma Essay Research Paper EmmasDilemma In

Emma?s Dilemma Essay, Research Paper Emma?s Dilemma In my investigation I am going to investigate the number of different arrangements of letters in a word. e.g. Tim Is one arrangementMit Is another First I am going to investigate how many different

Emma?s Dilemma Essay, Research Paper

Emma?s

Dilemma In my investigation I am going to investigate the number of

different arrangements of letters in a word. e.g. Tim Is one arrangementMit Is another First I am going to investigate how many different

arrangements in the name LUCY, which has no letters the same. LUCY LUYC LYCU LYUC LCYU LCUY ULCY UCLY UYLC ULYC UCYLThere are 4 different letters and there are? 24different arrangements.SAM SMA MSA MAS ASM AMSThere are 3 different letters in this name and 6 different

arrangements.JO OJThere are 2 different letters in this name and there are 2

different arrangements. UYCL????????????????????????? YCUL YUCL YULC YLCU YLUC CYLU CYUL CUYL CULY CLUY 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 AMME AMEM EMAM AEMM EAMM MMEA MMAE MEMA MAME MEAM MAEM4 letter word, 2 letters repeated, 12 different

arrangements.MUM MMU UMM3 letter word, 2 letters repeated, 3 different arrangements. PQMMM PMQMM PMMQM PMMMQ QPMMM QMPMM QMMPM QMMMP MPQMM MPMQM MPMMQ MQPMM MQMPM MQMMP MMPQM MMQMP MMMPQ MMMQP MMPQM MMMQP 5 letter word, 3 letters repeated, 20? different arrangements.SMMM MSMM MMSM MMMS4 letter word, 3 letters repeated, 4 different arrangements.RRRRK RRRKR RRKRR RKRRR KRRRR5 letter word, 4 letters repeated, 5 different arrangements.

Table of Results No of letters (n) No letters repeated Same letter repeated 2x (p) Same letter repeated 3x (p) Same letter repeated 4x (p) Same letter repeated 5x (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! ??????????????????????????????? ?(2×1 = 2) ???3 letters the same =??

n! ??????????????????????????????? ( 3×2x1 = 6)4 letters the same =??

n! ?????????????????????????????? ? (4×3x2×1 = 24)5 letters the same =??

n! ???????????????????????????????? (5×4x3×2x1 = 120)6 letters the same =??

n! ??????????????????????????? ?????(6×5x4×3x2×1 = 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, aaabb, or bbbaaa.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)xxyy???? xyxy???? yxxy xyyx???? yxyx???? yyxxThis is a 5 letter word xxxyy xxxyy?? xxyxy?? xxyxx?? xyxyx?? xyxxy xyyxx?? yyxxx?? yxxxy?? 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 factorised 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 : 1×2x3×4 ?????? (1×2) x (1×2) = 24 ??? 4 ???? =

6 different arrangementsA five letter word like aaaab; this has 4 a’s and 1 b (4 x’s

and 1 y) So: 1×2x3×4x5 ????? (1×2x3×4) x (1)

?????????? = 120 ?????????????? 24???? = 5

different arrangements ? A five letter words like abcde; this has 1 of each letter (no

letters the same) So : 1×2x3×4 ?? (1×1x1×1x1×1) ????????? = 24 ????????????? 1??????? = 24 different arrangementsA five letter word like aaabb; this has 3 a’s and 2 b’s (3

x’s and 2 y’s). So : 1×2x3×4x5 ?????? (1×2x3) x

(1×2) = 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 sameBy Katherine Bond? 11W

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