• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
• Level: GCSE
• Subject: Maths
• Word count: 1814

# Emma's Dilemma

Extracts from this document...

Introduction

Emma’s Dilemma

In this investigation, I will be attempting to find out the formula that would give me the number of possible arrangements for any group of letters, even if the group contains the same letters.

To find this out, I will be investigating the possible arrangements for different groups of letters and from there I will be then using the number of arrangements for each of these groups to find a formula.

I will start off with groups of different letters and find the formula for that.

The first group of letters I will start off with is the group ‘LUCY’:

1. LUCY
2. LUYC
3. LCUY
4. LCYU
5. LYCU
6. LYUC
7. ULCY
8. ULYC
9. UYCL
10. UYLC
11. UCLY
12. UCYL
13. CLUY
14. CLYU
15. CYLU
16. CYUL
17. CULY
18. CUYL
19. YLCU
20. YLUC
21. YULC
22. YUC L
23. YCUL
24. YCLU

Therefore for a group of four different letters there will be 24 different arrangements.

I will now attempt to find out the number of arrangements for a group of 3 different letters, ‘EMO’:

1. EMO
2. EOM
3. MOE
4. MEO
5. OEM
6. OME

So there are 6 possible different arrangements for any group of 3 different letters.

I am now going to try to find out the number of arrangements for a group of 2 different letters, ‘EM’:

1. EM
2. ME

Therefore, there are only two different arrangements for any two letter group of different letters.

I also

Middle

2

1

I tried my former formula out on these results but it did not work as I’d have to use the number of letters the same and the number of letter in the word both in the formula and in my formula I only have room for one number.

I will now find the number of combinations for a group of 3 letters with a group of the same letters in it. I will start off with a three letter word:

Two letters the same, ‘ITT’:

1. ITT
2. TIT
3. TTI

So there are only three different combinations for a three letter word with two of the same letters.

Three letters the same, ‘TTT’:

1. TTT

So there is only one different combination for a three letter word with three of the same letters.

Here is a table of a three letter word:

 Number of letters the same 1 2 3 Number of different combinations 6 3 1

The formula does not work for this one either; therefore I will have to find a different to my previous one however it might have something to do with the previous formula. It might involve ‘factorial’ as I had to use it in the previous formula but to find this overall formula out I will have to work out the number of different combinations for groups of different numbers with a group of the same number of letters in it.

I will now find the number of combinations for a group 4 of letters with a group of the same letters in it. I will start off with a four letter word:

Two letters the same, ‘EMMA’:

1. EMMA
2. EMAM
3. EAMM
4. MAME
5. MAEM
6. MEAM
7. MEMA
8. MMEA
9. MMAE
10. AMEM
11. AMME
12. AEMM

So there are only twelve different combinations for a four letter word with two of the same letters.

Three letters the same, ‘EMMM’:

1. EMMM
2. MEMM
3. MMEM
4. MMME

So there are only four different combinations for a four letter word with three of the same letters.

Four letters the same, ‘MMMM’:

1. MMMM

So there is only one different combination for a four letter word with four of the same letters.

Here is a table of a four letter word:

 Number of letters the same 1 2 3 4 Number of different combinations 24 12 4 1

Conclusion

Here are all the possible combinations for ‘AABBB’:

1. AABBB
2. ABABB
3. ABBAB
4. ABBBA
5. BABBA
6. BABAB
7. BAABB
8. BBABA
9. BBAAB
10. BBBAA

There are ten possible arrangements for a group of 5 letters, with a group of 3 letters the same and a group of 2 letters the same.

I will start off by using my former formula:

5!

3!

The answer to this is twenty. I have to find out someway to place the other number in the formula to get ten from twenty. Coincidentally, twenty divided by two, the other group of letters the same in the group of letters, is equal to the answer.

This would make my formula into this:

n = the number of letters in the word

y = the number of the letters the same

z = the other number of letters the same (if any)

x = the other number of letters the same (if any)        …etc

n!

y!z!x!

I believe this formula is correct but first I will have to try out again:

Example:

‘AABB’

1. AABB
2. ABAB
3. ABBA
4. BBAA
5. BABA
6. BAAB

The number of combinations for this example is 6. Now I have to try to see if it works with my formula:

4!

2! x 2!

The answer to this is the same as my result. This proves that my overall formula is correct and will work for any possible group of letters and will also get all of its possible combinations.

This student written piece of work is one of many that can be found in our GCSE Miscellaneous section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Miscellaneous essays

1. ## Mathematics Handling Data Coursework: How well can you estimate length?

1.9 5 0.1 50 1.9 ? l ? 2.0 1 0.1 10 2.0 ? l ? 2.1 2 0.1 20 Year 11 Frequency Density Table Length Frequency Class Width Frequency Density 1.0 ? l ? 1.1 1 0.1 10 1.1 ? l ? 1.2 1 0.1 10 1.2 ?

2. ## Data Handling

Many factors can be analysed here, was some people not relaxed, was there any miss calculations or miss readings when taking their pulses or time-sing by 4. PLAN During this investigation I will: 1. Select the grids that I need to do to analyse my hypothesis.

1. ## GCSE STATISTICS/Data Handling Coursework 2008

there are no height or weight values for students in year 11, so BMI cannot be calculated. I will take the number of students with height, weight and bleep test values for each year to use for the sample. This table shows the number of pupils with all three values for each year.

2. ## Data Handling

5 Age Age is very important figure, which make it very essential, and a key figure when purchasing the car. Age is another considerable figure and surely will affect second hand car price. I predict that as the age increases, the second hand car price will decrease.

1. ## MATHS COURSEWORK - Mayfield High - To analyse data provided by Mayfield High School ...

Bias is anything which distorts the data so that the end result is it cannot provide a fairly representative picture of a population. Bias can arise in two different methods: In the way the questions were asked and from the people asked.

2. ## Maths Data handling Corsework

For the second hypothesis, I will use a set of box plots, comparing the different pieces of data. Finally for the third hypothesis I will use cumulative frequency to compare two sets of data presented by curves. Then I will analyse and interpret each one carefully, to see whether my hypotheses were correct or not and why.

1. ## Borders - find out the differences in the patterns of the colored squares.

The following is an example to test if I am right: 4N = number of white squares N = 10 4 x 10 = 40 This gives me the correct amount of white squares; this is true because of the table of new orders gives me this answer.

2. ## Statistics coursework. My first hypothesis is that people with a smaller hand span ...

The UQ is 15.35 Box and whisker diagrams: A good way of showing overall results is to use box and whisker diagrams: This shows that overall males have a slightly faster reaction time than females. This proves my hypothesis is right.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to