• 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

# Emma&amp;#146;s Dilemma

Extracts from this document...

Introduction

Denise Shaw

Maths Coursework

## Emma’s Dilemma

I am using the name Emma playing with the letters in the name to make different arrangements. Emma is a 4 letter word I should get 24 different ways of writing it but because it has 2 letters the same I will only get 12 different ways of writing it.

The ways are:-

1.emma     7.mmae

2.eamm     8.mema

3.emam     9.mame

4.meam    10.aemm

5.maem    11.amme

6.mmea    12.amem

I am now going to use the word Lucy, Lucy also has 4 letters in it but because

Middle

3.mto   6.omt

Now I am going to use AA a 2 letter word with 2 letters the same, I will get 1 arrangement.

The 1 arrangement is:-

1.aa

And Jo a 2 letter word with no letters the same gives 2 different ways.

The 2 different ways are:-

1.jo

2.oj

 #letters in name 0 letters repeating 1 letter repeating 2 2 1 3 6 3 4 24 12

I predict that if I do a 5 letter word with 2 letters the

Conclusion

10.dnnya   20.anynd   30.nadny     40.nnday    50.yandn    60.ynnda

I predict for a 6 letter word with no letters the same I would get 720 different arrangements. The way to do this is start from 1 and go all the way up to 6 and times them all together and you get the answer for example 1X2X3X4X5X6 =720 arrangements.

To find out the answer on a calculator the formula is :-

 n shift x-1 =

## This is called factorial

Factorial is the product of all the positive ategers from 1 to a given number for example 4 factorial, usually written 4! Is product 1.2.3.4=24

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma 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 Emma's Dilemma essays

1. ## Emma's Dilemma

X 2!) 5 30 = 5! / (2! X 2!) Table 5, Two letters repeated three times: Number of Letters: Number of different combinations: 6 20 = 6! / (3! X 3!) Rule: From these sets of results, I can see that if I expand the rule I found to solve "Two letters repeated twice", then this rule

2. ## GCSE maths coursework: Emma's dilemma

10 arrangement so on ------10 arrangements 112212 121122 112221 122112 Total arrangement= 20 the formula work I expect the arrangements for 8 fig will be a= (1*2*3*4*5*6*7*8)/1*2*3*4*1*2*3*4=70 let's confirm 11112222 11211222 11221212 11121222 11212122 11221221 11122122 11212212 11222112 --------- 15 arrangements 11122212 11212221 11222121 11122221 11221122 11222211 12111222 12122112 12212121

1. ## I have been given a problem entitled 'Emma's Dilemma' and I was given the ...

/ 2! = No. of Permutations and 4! / 2! = 12 and not 6 - which is the correct answer. This is interesting however as the correct answer is half the number that my formula produced. So I thought about this when looking at the further variations.

2. ## Emma's Dilemma

First, I will investigate a 3-letter word. CCC There is only 1 arrangement. Now I am going to investigate a 4-letter word. AAAR AARA ARAA RAAA There are 4 arrangements. Now I will investigate a 5-letter word. YYYEL YYYLE YYEYL YYELY YYLEY YYLYE YEYYL YEYLY YELYY YLYYE YLYEY YLEYY EYYYL

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