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  • Level: GCSE
  • Subject: Maths
  • Word count: 3404

Emma’s Dilemma

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Year 10 Maths Coursework Emma's Dilemma Emma is playing with different arrangements of her name. One arrangement is EMMA A different arrangement is MEAM Another arrangement is AEMM I investigated the total number of different arrangements of the letters of Emma's name. I worked out all the arrangements and come up with the following EMMA MAME MMAE AMME EMAM MAEM MAEM AMEM EAMM MEAM MAEM AEMM From this I can see that there are 12 possible arrangements. It was also interesting to see that when starting with each different letter in Emma's name, there was the same amount of arrangements for each. We had to account double the M's with there being two M's in Emma's name and so their was two times as many arrangements for that letter. Emma has a friend named Lucy. When investigating the number of different arrangements of the letters of Lucy's name, I came up with the following arrangements. LUCY CULY ULCY YUCL LCUY CLUY UCLY YCUL LYUC CYUL UYLC YLUC LYCU CYLU UYCL YLCU LCYU CLYU UCYL YCLU LUYC CUYL ULYC YULC The total number of different arrangements for the letters in Lucy's name is 24. It is interesting to see that even though the two names both have the same amount of letters in them, the number of arrangements of the letters comes to different totals. We can see that this is due to the double M in Emma, meaning that some arrangements would not be counted as we would merle swap the two M's which does not count as an alternative arrangement. We can also see how each letter at the start has the same number of arrangements. By seeing that this is the case in both Emma's and Lucy's names it is important to make sure that this rule applies to all other names. To show that this rule works, I checked it with a three-letter name. ...read more.


The double letter merle means that there is two of the double letter. Emma shows the arrangements for a four-letter name and I also looked at the five-letter name of Kitty. I looked into the arrangements beginning with the letter K and then was able to use my rule. KITTY KTITY KTYIT KYTTI KITYT KTTIY KTTYI KYITI KIYTT KTIYT KTYIT KYITT If we put this into the rule L x number of arrangements for one letter = A 5 x 12 = 60 We can then put the information for double lettered words into a table to display the information more clearly. No of letters No of arrangements No of arrangements (double letters) (all letters the same) 1 - 1 2 1 2 3 3 6 4 12 24 5 60 120 We can see how the same general patterns occur in the table as seen with all letters different. The increase between two numbers is shown next to the higher number. We can see the general trend that as the number of letters goes up so does the number of arrangements. When drawing out the table for names with all letters different, I included a column showing the increase between each number of arrangements. I decided not to include this in this column, as it did not play any important role. I have included the information about names with all letters the same so that I can easily compare the two and look for differences. When looking at the information in the table, it is easy to see how we are able to use the same rule Number of letters (L) x previous number of arrangements (PA) = number of arrangements (A) e.g 4 x 3 = 12 It is also important to note how we can compare double lettered words with words with all the letters different. We can see that double lettered words have half the number of combinations as single lettered words. ...read more.


L x number of arrangements for one letter = A 7 x 5 = 35 This then gives us the correct total, proving the formula to work. 2. YXYYX L! / SL! = A 5! / 3! X 2! = 10 Arrangements: XXYYY XYYYX YYYXX YYXYX YYXXY XYYXY XYXYY YXYYX YXYXY YXXYY Again this shows that the formula works by showing ten arrangements. Although on the previous test, I only wrote out a few arrangements and used the formula to complete the rest, I did not for this. This is because there are not as many for this example, and I feel safer by righting them out if it is possible. However, when there are a lot, it is not practical to write them all and so that is why I use the formula. 3. YYYXXYXY L! / SL! = A 8! / 5! X 3! = 56 For this example it will be easier to work out the arrangements for one letter and work out the rest with the formula. This means that our total number needs to be divided by the number of letters. 56 / 8 = 7. I will use this to find out all the Xs, but first this number needs to be multiplied by however many Xs there are. 7 X 3 = 21. This means that I need to find 21 arrangements beginning with X. Arrangements XXXYYYYY XYYYYXYX XYYXYXYY XXYYYYYX XYYYYXYX XYYXYXYY XXYYYYXY XYYYXYYX XYXYYYYX XXYYYXYY XYYYXYXY XYXYYYXY XXYYXYYY XYYYXXYY XYXYYXYY XXYXYYYY XYYXYYYX XYXYXYYY XYYYYYXX XYYXYYXY XYXXYYYY This shows 21 arrangements all beginning with X. It means that there is 7 arrangements for each letters. I will put this in my formula. L x number of arrangements for one letter = A 8 x 7 = 56 This again proves that the formula works for any combination. By this I can say that the formula Number of letters! (L!) / number of same letters! (SL!) = Arrangements (A) Will give the correct number of arrangements for any given word or name. Harriet Kemp 1 1 ...read more.

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