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  • Level: GCSE
  • Subject: Maths
  • Word count: 1728

Dave's Dillemma.

Extracts from this document...

Introduction

Dave's Dillemma

Dave is playing with arrangements of the letters of his name.

Here are the arrangements for DAVE:



As you can see, there are 24 possible arrangements for the name DAVE. There are four letters in the name, and six arrangements starting with each letter.

4! = 4 x 3 x 2 x 1

=24

(! Means factorial and is a button found on a scientific calculator)

e.g. 4! = 4 x 3 x 2 x 1 = 24

3! = 3 x 2 x 1 = 6


Here are the arrangements for
DAVE's friend EMMA:


There are 12 possible arrangements for the name EMMA. There are four letters in the name, but one is repeated twice. Because of this we get only 12 arrangements opposed to 24 and this is because the two letters are repeated twice.


Here are the arrangements for
EMMM:


There are 4 arrangements for EMMM, one-third of the arrangements for EMMA, which are 12.

This tells us that the more frequently that the letters appear in a name, the fewer arrangements there will be for that name.

Here is a table showing this (The number of letters-4 will be kept constant):



DAVE has 24 arrangements because none of the letters are the same.

EMMA has 12 arrangements because 2 of the letters are the same.

EMMM has 4 arrangements because 3 of the letters are the same.

MMMM has 1 arrangement because all of the letters are the same.

...read more.

Middle


I have found a pattern from the arrangements:


As you can see there is a pattern forming. The number of letters multiplies the previous number of arrangements each time.

e.g. -no. of letters = 4

no. of arrangements = 24 (previous number of arrangements)

-no. of letters = 5 (number of letters in this case)

Here are some different arrangements in terms of x's and y's:

Arrangements for xxyy:



As you can see there are 2 x's and 2 y's. There are 6 arrangements, 3 starting with each letter.

Arrangements for xxxyy:



There are 10 arrangements. There are 6 starting with x and 4 starting with y.

Arrangements for xxxxy:



There are only 5 arrangements. There are 4 starting with x and 1 starting with y.

Because there are two sets of same letters the previous formula cannot be used. However I have come up with another formula:

x!

(y! x z!)

x = total number of letters, y = number of repeated letters (1), and

z = number of repeated letters (2).

If there were an arrangement of 3 x's and 3 y's, i.e. xxxyyy, and we separate the x's from the y's, so that we have (x + y). (x + y) is the total number of letters. When re-arranging, the first letter that we would chose would be (x + y), the second (x + y - 1) because one letter has already been chosen, and so on.

...read more.

Conclusion

(x + y + z) (x + y + z - 1) (x + y + z - 2) (x + y + z - 3) (x + y + z - 4)

(x) (y) (y - 1) (z) (z - 1)

because it has 5 letters, 3 of which are different.

(1 + 2 + 2) (1 + 2 + 2 - 1) (1 + 2 + 2 - 2) (1 + 2 + 2 - 3) (1 + 2 + 2 - 4)

(1) (2) (2 - 1) (2) (2 - 1)

= 5 x 4 x 3 x 2 x 1

1 x 2 x 1 x 2 x 1

= 5!

1! x 2! x 2!

= 120

4

= 30

This proves that my formula works, and in future the 1! Does not need to be included because in names like THANCANAMOOTOO in which there are 3 letters that are not repeated (H, C, M) we do not need to write

1! x 1! x 1! Because 1! Is equal to simply 1 and this does not have any effect on the formula if we exclude it.

The formula can be simplified to

x!

y! x z! x a! …and so on, depending on the number of letters which are different.

x is the number of letters

y is the number of letters repeated, as are z and a.

This formula can go on forever depending on the number of letters repeated twice or more.

So, for example if in a name like THANCANAMOOTOO we would use the formula to work it out like this:

14!

2! x 3! x 2! x 4!

= 14!

2 x 6 x 2 x 24

= 151,351,200

· It is 14! Because that is the number of letters in the name.

· It is divided by (2! x 3! x 2! x 4!) because the letters T and N are both repeated twice, the letter A is repeated 3 times, and the letter O is repeated 4 times, and hence we multiply the factorials of them by each other.

...read more.

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