27th June 2001 Tom Pountain 10AEmma's DilemmaI investigated the number of different arrangements of four letters with no repetitions.1)ABCD2)ABDC 3)ADBC4)ADCB 5)ACBD 6)ACDB 7)BACD 8)BCAD 9)BDCA 10)BADC 11)BDAC12)CDBA 13)CBAD15)CDAB16)CADB17)CBDA18)DABC19)DBCA20)DCAB21)DACB22)DBAC23)DCBA24)DABC25)BCDA I have found 24 different arrangements of these letters and this result is confirmed in the tree diagram. Secondly, I have investigated the number of different arrangements of four letters with one letter repeated twice.1)ACBB 2)ABCB 3)BABC 4)BACB 5)BBAC 6)BBCA 7)BCAB8)BCBA9)CBBA10)CABB11)CBAB12)ABBCI have found 12 different arrangements of the letters and this result is confirmed in the tree diagram. From these two investigations, I have worked out a method that can be used for further work: Firstly, with ABCD you rotate the last two letters, then you get ABDC. Then, with ABCD you must then rotate the last three letters and try the possibilities of ADBC, ADCB, ACDB, ACBD. Because the letter 'B' has been the first number of last three letters before, we don’t do it again.We have list all arrangements with A first, so now we do B secondly: BACD, and we do same thing to it, it will like this:BACD=BADC, BADC=BDCA...BDAC...BCAD...BCDAWe have finished with B first, then we do C next:CABD=CADB, CADB=CBDA...CBAD...CDAB...CDBAWe have finished with C first, then we do DDABC=DACB, DACB=DBCA...DBAC...DCAB...DCBAWe have listed all of the arrangements of ABCD, so with this method we can now use this method to arrange 5 letters, or more.1)ABCD2)ABDC 3)ADBC4)ADCB 5)ACBD 6)ACDB 7)BACD 8)BCAD 9)BCDA10)BDCA 11)BADC 12)BDAC13)CDBA 14)CBAD15)CDAB16)CADB17)CBDA18)CABD19)DBCA20)DCAB21)DACB22)DBAC23)DCBA24)DABCLooking at each number at the beginning of the arrangement, if we times 6 by 4, we would get 24, and we see the total arrangements of 24.Let’s now try this with 5 letters and no repetitions:ABCDEABCEDABECDABEDCABDECABDCEACBDEACBEDACEBDACEDBACDEBACDBEADBCEADBECADEBCADECBADCEBADCBEAEBCDAEBDCAEDBCAEDCBAECDBAECBDThe arrangements of the last 4 letters with 'A' first added up to 24 again, so if we timed 24 by 5 we would get 120, and 120 is also the total of arrangements of 5 letters according to the tree diagram. Then if there are 6 letters in an arrangement, then the total amounts of arrangements should be 120 times by 6, or 720, and according to the tree diagram 720 is the total of arrangements. So if an arrangement has 7 letters, then the total number of arrangements should be 720 times by 7, or 5040, so the total amount of arrangements would be 5040.This is my prediction, so now I'll try and find a formula to show this. I can begin by looking at how the amount of arrangements is mathematically formed for each number of letters that I have looked at.4 4x65 4x6x56 4x6x5x67 4x6x5x6x78 4x6x5x6x7x8Since this only begins with '4x6' each time because I originally investigated four number arrangements, there should be a more rounded way of presenting this data. In retrospect, I can see that the '4x6' can be replaced with '1x2x3x4' to the same affect. 4x6=241x2x3x4=24So now, I can rewrite that
prediction in a better and universal way.n1 = 1n2 = 1x2n3 = 1x2x3n4 = 1x2x3x4n5 = 1x2x3x4x5n6 = 1x2x3x4x5x6From my own mathematical knowledge, I can see that the data above can be represented by the factorial, '!', of each number. This is because: 1! = 12! = 1x23! = 1x2x34! = 1x2x3x45! = 1x2x3x4x56! = 1x2x3x4x5x6So if we use 'n' to represent the number of letters in an arrangement, then it has a total amount of arrangements of n!.The formula: a = n!n! can easily be calculated on calculator. You must first press key 'n' (whatever the number of letters ...
This is a preview of the whole essay
prediction in a better and universal way.n1 = 1n2 = 1x2n3 = 1x2x3n4 = 1x2x3x4n5 = 1x2x3x4x5n6 = 1x2x3x4x5x6From my own mathematical knowledge, I can see that the data above can be represented by the factorial, '!', of each number. This is because: 1! = 12! = 1x23! = 1x2x34! = 1x2x3x45! = 1x2x3x4x56! = 1x2x3x4x5x6So if we use 'n' to represent the number of letters in an arrangement, then it has a total amount of arrangements of n!.The formula: a = n!n! can easily be calculated on calculator. You must first press key 'n' (whatever the number of letters is), then press key '!', then you can find the total amount of arrangements.The formula can be confirmed when I look at the following data.1! = 1 = 12! = 1x2 = 23! = 1x2x3 = 6 4! = 1x2x3x4 = 245! = 1x2x3x4x5 = 1206! = 1x2x3x4x5x6 = 720I believe that the reason for this formula to be n! Is explained when you imagine that every letter of an arrangement has a space in a grid, such as ABCD:At first, you would begin with a blank grid, like thisThen, firstly with the A, you would have a possibility of 4 spaces for it.Secondly, with the B, there only be 3 possibilitiesFor the same reason, there would only be 2 possibilities for C and 1 for D.So to find the overall probability of ABCD's possible arrangements, you must multiply 4 for A, 3 for B, 2 for C and 1 for D.4x3x2x1 = 12 = 4!The possibility is also shown on the tree diagrams.I now have to investigation arrangements that have a repetition of two of the same figures, for example: AAB, ABBC.I have investigated the arrangements of AAB:AABABABAAI have found 3 different arrangements of these letters.I have investigated the arrangements of ABBC:1.ABBC 2.ABCB 3.ACBB 4.BBCA5.BCAB6.BCBA7.BABC8.BACB9.BBAC 10.CABB11.CBAB 12.CBBAI have found 12 different arrangements of these letters. From the last investigation, I have decided to look at the last results, in this case '3' for a pattern. In relation to '3', 12 = 3x4, 4 being the number of letters in the arrangement.I have investigated the arrangements of ABBCD:1 122342 122433 12324 4 123425 124236 12432 7 132248 132429 1342210 1422311 1423212 1432212x5 = 60I have found 60 (12x5) different arrangements of these letters.Now I can show alternative data, after learning from the last investigation, for an arrangement with one repetition of 2 of the same letters.3 = 1x34 = 1x3x45 = 1x3x4x5They are not multiplied by 2 because 2 of the numbers (those repeated) can only be used once each in a certain space. I can now try and find a formula in terms of n!.if n = number of figures, and a= number of arrangementsthe formula is: a = n!/2This is because the lack of 'x2' in the equation acts the same as '/2' so the formula would be just n! With that adjustment. This can be seen when looking at the tree diagrams. The formula can be confirmed when I look at the following data.2 (1x2)/2 = 1 3 (1x2x3)/2 = 3 4 (1x2x3x4)/2 = 12 The reason for this formula is that it would be n!, for the same reason as aforementioned with:But now, because of the two B's, there would be some arrangements from the tree diagram that are the same. I.e., ABBA and ABBA. Because of their being two of each arrangement, the total amount for four non-repeated letters, 24, should be halved. So the formula would be n! divided by 2. I now have to investigation arrangements that have a repetition of three of the same figures, for example: AAA, BBBC.I have investigated the arrangements of AAA:AAAI have just found 1 arrangement of these letters.I have investigated the arrangements of BBBC:33313313 31331333Try 3331233312 31233 12333 33321 31323 13233 33123 31332 1332333132 32331 1333233231 32313 33213 3213321333231332331323331I have found 20 arrangements of these letters.I have investigated the arrangements of AAABCD:333124 332134333142 332143 333214 334321333241 332314333412 332341 333421 332413331234 332431331243 334123331324 334132331342 334213331423 334231331432 334312312334 321334312343 312433 34....313234313243 313324 313342313423313432314233314323314332123334 133324 123343 133342 123433 133234124333 133243124332 133423 134323 133432 134233 142333 132334 143233132343 143323132433 143332 I have found 120 arrangements of these letters. Looking back at previous results, in terms of 6, the total arrangements of an arrangement that have a repetition of three of the same figures is 20x6 = 120Now I can show the data for an arrangement with one repetition of 3 of the same letters.3 14 1x45 1x4x56 1x4x5x6They are not multiplied by 2 nor 3 because 3 of the numbers (those repeated) can only be used once each in a certain space. I can now try and find another formula in terms of n!.if n = number of figures, and a= number of arrangementsthe formula is: a = n!/6This is because the lack of 'x2' and 'x3' in the equation acts the same as '/(2x3)'='/6' so the formula would be just n! With that adjustment.This formula is such for the same reason as n!/2, but this time it must not only be halved, but divided by 3 also, this can by simplified :2x3 = 6So the simple formula is n!/6. Now that we have found formulae for arrangements with both 2 and 3 repeats respectively, we can now review the formulae and find a universal formula any number of repeats.Formula for different numbers in arrangement: a = n!Formula for 2 same numbers in arrangement: a = n!/2Formula for 3 same numbers in arrangement: a = n!/6We can look at the formulae in a table:n123x126n represents the number of figures of a number, and X represents the divided number in the formula.From looking at the table, I can see that x equals the number of repetitions factorised. x=r!In the table, x (r!) equals the last x (r!), times n, so I expect the formula for 4 same numbers in arrangement is a= n!/24I investigated the number of different arrangements of four letters with four repetitionsAAAAI have found 1 different arrangement of these lettersHere is the formula:a=4!/24=(1x2x3x4)/24=1 I investigated the number of different arrangements of five letters with five repetitions1 AAAAB2 AAABA3 AABAA 4 ABAAA5 BAAAAI have found 5 different arrangements of these letters.Here is the formula:a = 5!/24 = (1x2x3x4x5)/24 = 5 I will investigate this formula, and try to improve it, with another table.n12345r!12624120Looking at this with my knowledge from this investigation, I can see a formula so if a represents arrangement, and n represent numbers of figures, r! represents the number of repetitions factorised, and the formula is: a = n!/r! But of course the '!' cannot be cancelled out because it is part of a multiplication.The reason for this formula is that the number of proper possibilities would be fewer than usual because some of the numbers are the same. But if we make another grid just for the repeats:There would always be just one possibility. So the formula of n! would have to divided by r! Because the repeats would be factorised themselves in the same as the total amount of letters in the arrangement.I now have to investigate arrangements that have two repetitions of the same letters, for example: AABB, AAABBB.I investigated the number of different arrangements of four letters with two repetitions of the same letters twice.1. 1122 2. 1212 3. 12214. 21225. 2212 6. 2221 I have found 6 different arrangements of these lettersHere is the formula:a = (1x2x3x4)/2 = 12The formula doesn't work now, as it needs to be halved.I investigated the number of different arrangements of six letters with two repetitions of the same letters thrice.111222 121212 222111 211212112122 121122 221211 211221112212 122112 221121 212112 112221 122121 221112 212121121221 122211 211122 212211 I have found 20 different arrangements of these letters.Here is the formula:a = (1x2x3x4x5x6)/(1x2x3) = 120 It doesn’t work but if we divide it by 6 again, or (1x2x3)the formula still works if we multiply the number of repetitions factorised again.For example:For an arrangements of four letters with two repetitions of the same letters twice:a = (1x2x3x4)/(1x2x1x2) = 6 = n!/(r!xr!) I expect this will still work for arrangements of six letters with two repetitions of the same letters thrice.To follow this formula, I predict that the arrangements for this isA = (1x2x3x4x5x6)/(1x2x3)x(1x2x3) = 20I investigated the number of different arrangements of six letters with two repetitions of the same letters thrice for the second time.111222 121221 122121 112122 121212 122211 112212 121122112221 122112I have found 20 different arrangements of these letters and the formula does work.I predict the arrangements for 8 letters will be: a = (1x2x3x4x5x6x7x8)/(1x2x3x4)x(1x2x3x4) = 70I investigated the number of different arrangements of eight letters with two repetitions of the same letters four times.11112222 11211222 11221212 11121222 11212122 1122122111122122 11212212 11222112 11122212 11212221 1122212111122221 11221122 1122221112111222 12122112 12212121 1212121212112122 12122121 12212211 1221121212112212 12122211 12221112 12112221 12211122 1222112112121122 12211221 1222121112121221 12212112 1222211135x2 = 70I have found 70 different arrangements of these lettersThis formula can be written as: a = n!/r!xr!The explanation for this formula is the same as n!/r! but if there are more repetitions then they have to be accounted for in the same way and multiplying the number of repetitions factorised by one another would have the same affect.I now have to investigate arrangements that have multiple repetitions of the same letters, for example: AAABB, AAAABB.I investigated the number of different arrangements of four letters with two repetitions of the same letters twice.I can to change the formula a = n!/r!xr!, because there are 2 pairs of the same numbers with different number of letters. So I could change the formula to a = n!/r1!xr2!. But this is mathematically similar.I investigated the number of different arrangements of five letters with two a repetition of the one letter twice and another thrice.According to the formula, I predict the total arrangement for this is:a = (1x2x3x4x5)/(3x2x1)x(2x1) = 1011122 1221111212 2111211221 21121 12112 2121112121 22111I have found 10 different arrangements of these lettersLet’s try 7 fig, with 3 same number, and 4 same number.I predict the total arrangement is: a = (1x2x3x4x5x6x7)/(3x2x1)x(4x3x2x1)=351112222 1222211 2222111 22112121121222 1222121 2221211 22112211122122 1222112 2221121 22121121122212 1221221 2221112 2212121 1122221 1221212 2211122 22122111212221 12211221212212121212212112222111222 2112221 2121221 21222112112122 2121122 2122112 2112212 2121212 2122121I have found 35 different arrangements of these letters.I now have to investigate arrangements that have three repetitions of the same letters, for example: AABBCC.The formula needs to be rewritten as a = n!/r!xr!xr! now. There are three r!'s , because there are now three pairs of repetitions. If there are two pair of same number of figures of same number, then there are only two xi need to multiply, and if there are two pair of different number of figures of same number, then there would be x1i and x2i need to multiply.Let’s confirm the formula.let’s try 112233, according the formula, I expect the total arrangements is a=(1x2x3x4x5x6)/(1x2)x(1x2)x(1x2)=90Let’s confirmed112233 121233 123123 131223 132231 112323 121323 123132 131232 132123112332 121332 123213 131322 132132 ------ 30 arrangements113223 122133 123231 132321 133122113232 122313 123312 132312 133212113322 122331 123321 132213 1332212------- 3------so on ------30 arrangements so on --------- 30arrangementsThe total arrangement is 90, the formula works.Formula is confirmedWhat about three pairs of different number of figures of a numberFor example:122333according the formula, the total arrangement is a=(1*2*3*4*5*6)/(1*1*2*1*2*3)=60Let’s confirm it:122333 212333 231332 3--------123233 213233 232133 so on ------- 30 arrangements123323 213323 232313 123332 213332 232331 -----30 arrangements 132233 221333 233123132323 223133 233132132332 223313 133213133223 223331 233231133232 231233 233312133322 231323 233321The formula worksFormula is confirmedFrom the investigation above we find out the formula for calculating the number of arrangements, it’s a=ni/xia represent the total arrangementsn represent the number of figures of the numberI represent the key Ix represent the numbers of figures of same number of the numberif there are more than one pair of same number, x2, or x3, so on may added to the formula, it depend how many pairs of same number.For example:for 2 pairs of same number of figures of same number of a numberthe formula is a=ni/xixifor 2 pairs of different number of figures of same number of a numberthe formula is a=ni/x1ix2ifor 3 pairs of same number of figures of same number of a numberthe formula is a=ni/xixixiform 3 pairs of different number of figures of same number of a numberthe formula is a=ni/x1ix2ix3i.The formula can be also used to the arrangements of letter.For example:xxyythe arrangement for this is a=(4x3x2x1)/(2x1x2x1)=6xxyyythe arrangement for this is a=(5x4x3x2x1)/(3x2x1)x(2x1)=10xxxxxxyyyyyyyyyythe arrangement for this is a=n!/x1!x2!=(16x15x14x13x12x11x10x9x8x7x6x5x4x3x2x1)/(10x9x8x7x6x5x4x3x2x1)x(6x5x4x3x2x1)=8008The total arrangement is 8008.Use this formula, we can find out the total arrangements of all numbers and letters.
