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• Level: GCSE
• Subject: Maths
• Word count: 1354

# Emma&amp;#146;s Dilemma

Extracts from this document...

Introduction

John Palfrey

May 1, 2007

John Palfrey, 5UL

Emma’s Dilemma

1)        Rearrangement Of EMMA

If we represent the letters in Emma’s name as numbers, we get an arrangement like this:

E        M        M        A

1        2        2        3

We can then use this system to find all of the other potential arrangements:

1        2        2        3

1        2        3        2

1        3        2        2

2        1        2        3

2        1        3        2

2        2        1        3

2        2        3        1

2        3        1        2

2        3        2        1

3        1        2        2

3        2        1        2

3        2        2        1

As you can see, there are 12 combinations.

2)        Rearrangement Of LUCY

If we represent the letters in Lucy’s name as numbers, we get an arrangement like this:

L        U        C        Y

1        2        3        4

We can then use this system to find all of the other potential arrangements:

1        2        3        4        2        1        3        4

1        2        4        3        2        1        4        3

1        3        2        4        2        3        1        4

1        3        4        2        2        3        4        1

1        4        2        3        2        4        1        3

1        4        3        2        2        4        3        1

3        1        2        4        4        1        2        3

3        1        4        2        4        1        3        2

3        2        1        4        4        2        1        3

3        2        4        1        4        2        3        1

3        4        1        2        4        3        1        2

3        4        2        1        4        3        2        1

As you can see, there are only 24 potential arrangements; twice as many as when there is one repetition.

3)        Other Rearrangements

If we use the name MAY to investigate the number of potential arrangements when there are only three factors, we get these arrangements:

M        A        Y

1        2        3

1        2        3

1        3        2

2        1        3

2        3        1

3        1        2

3        2        1

We only get six arrangements, a quarter of the amount when we use four factors.

Middle

1        2        1        3        2        3        1        1

1        2        3        1        3        1        1        2

1        3        1        2        3        1        2        1

1        3        2        1        3        2        1        1

2 factors, 1 repetition        =        1 arrangements        =        (1 x 2) / 2        =        2!

2

3 factors, 1 repetition        =        3 arrangements        =        (1 x 2 x 3) / 2        =        3!

2

4 factors, 1 repetition        =        12 arrangements        =        (1 x 2 x 3 x 4) / 2        =        4!

2

If we investigate the arrangements when there are three repetitions, we get these results:

1        1        1

1        1        1        2

1        1        2        1

1        2        1        1

2        1        1        1

1        1        1        2        3        1        3        1        2        1

1        1        1        3        2        1        3        2        1        1

1        1        2        1        3        2        1        1        1        3

1        1        2        3        1        2        1        1        3        1

1        1        3        1        2        2        1        3        1        1

1        1        3        2        1        2        3        1        1        1

1        2        1        1        3        3        1        1        1        2

1        2        1        3        1        3        1        1        2        1

1        2        3        1        1        3        1        2        1        1

1        3        1        1        2        3        2        1        1        1

3 factors, 2 repetitions        =        1 arrangements        =        (1 x 2) / 2        =        3!

6

4 factors, 2 repetitions        =        4 arrangements        =        (1 x 2 x 3) / 2        =        4!

6

5 factors, 2 repetitions        =        20 arrangements        =        (1 x 2 x 3 x 4) / 2        =        5!

6

There is a pattern in the denominatives, in that they are equal to the factorial of the number of repeated factors. In the sequence 1112, there are three repetitions, one number repeated twice. Therefore, the formula for this sequence would be:

4th         =        x! / r!

=        4! / 3!

=        24 / 6

=        4

and there are indeed 4 arrangements. In the sequence 11123, which also has three repetitions, the formula would be:

5th         =        x! / r!

=        5! / 3!

=        120 / 6

=        20

this is also the right amount. This also works for when there are two repetitions:

4th

Conclusion

x        z        y        y        x        z        x        y        y        x

y        x        x        y        z        z        y        x        x        y

y        x        x        z        y        z        y        x        y        x

y        x        y        x        z        z        y        y        x        x

then the formula, if rearranged, should be

5th         =        x! / (r1! x r2!)

=        (5! / (2! x 2!)

=        120 / 4

=        30

so this formula is right. If we take it one step further and account for each letter, we see that;

5th         =        x! / (r1! x r2! x r3!)

=        (5! / (2! x 2! x 1!)

=        120 / 4

=        30

it doesn’t make much difference, as multiplying by 1 never has any effect anyway. If we test this formula on the first and second sequences (4 letters, no repetitions and 4 letters, 1 repetition);

4th         =        x! / (r1! x r2! x r3! x r4!)

=        4! / (1! x 1! x 1! x 1!)

=        24

4th         =        x! / (r1! x r2! x r3! x r4!)

=        4! / (2! x 1! x 1! x 1!)

=        24 / 2

=        12

The formula even works for these two, so it is safe to assume that it works for all of the combinations of every set of letters with any number of repetitions.

nth         =        n! / (r1! x r2! x r3! x r4!...)        ;        r1 + r2 + r3 + r4...        =        n

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

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