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Introduction

## 1) K

This is because there is only 1 space for the letter to go.

1

With 2 letters in the word there are 2 different arrangements.

1) JO 2) OJ

This is because there are 2 choices for the first letter and 1 for the last.

2 1

With 3 letters in the word there are 6 different arrangements.

1) LIZ 2) LZI 3) ILZ 4) IZL 5) ZLI 6) ZIL

This is because there are 3 choices for the first letter, 2 for the next and then 1

3 2 1

 No. of letters Working Arrangement 1 1 1 2 2x1 2 3 3x2x1 6

From this initial assessment the rule for any size word combination, where all the letters are different, is a= n!  where n is the number of letters in the word and a is the total number of arrangements.

I would predict, therefore, that there will be 4! or 24 arrangements in the word Lucy.

Middle

 Words with different letters Words with 2 letters the same Words with 3 letters the same No. of letters No. Arrangements No. Arrangements No. Arrangements 3 6 3 1 4 24 12 4 5 120 60 20 6 720 360 120

With three letters the same (AAA) there is only one arrangement that is seen. However if they were labelled A1A2A3 we would see that there are in fact six (or 3!) combinations in practise. Consequently the equation becomes

a = n! / 3!

Where n = the number of letters in the word and a = the number of different arrangements.

To confirm the prediction I will look at a six letter word with one letter repeated three times. If the formula is correct there will be 6! / 3! different arrangements (i.e.120)

Assume the word is ABCDDD

If A B or C are the first letter there will be 3 x 20 arrangements as the last four letters will include one letter, D, three times. This is like AONNN where I have already shown there are 20 arrangements.

If D is the first letter there will be 5! / 2! arrangements (i.e. 60) as the last five letters include one, D, that is there twice ( see table above)

This gives a total of 120 different arrangements as predicted.

I therefore conclude that the connecting formula is

a= N! / r!

Where n = the number of letters in the word, r is the number of letters repeated and a = the number of different arrangements.

 No. of letters Words with different letters Words with 2 letters the same Words with 3 letters the same Words with 4 letters the same No. Arrangements No. Arrangements No. Arrangements No. Arrangements 4 24 12 4 1 5 120 60 20 5 6 720 360 120 30

Conclusion

is the number of letters in the word, x y z etc are the number of different number combinations and a is the number of arrangements.  It should be noted that  n  =  x + y + z …….

Final Conclusion

From the work I have done I have developed a number of formula, some of which are for specific circumstances, leading to the universal formula which covers all possibilities.

1. For words containing all different letters I have

a = n!

Where n = the number of letters in the word and a = the number of different arrangements.

This can also be written a = n! / 1! x 1! x 1!……….

1. For words containing a letter repeated twice the equation is

a =n! / 2!

Where n = the number of letters in the word and a = the number of different arrangements.

This can also be written a = n! / 2! x 1! x 1! x 1!……

1. For words containing a letter repeated three times the equation is

a = n! / 3!

Where n = the number of letters in the word, and a = the number of different arrangements.

This can be written as a = n! / 3! x 1! x 1! x 1!…..

The universal formula, covering all possibilities, is

a = n! / x! x y! x z!…….

Where n = the number of letters in the word, x y z…. are the number of similar individual letters and a is the number of arrangements.

n will be the sum of x + y+ z +….

It is seen that the formula 1), 2) and 3) fit the universal formula.

From this formula a master table can be achieved for any number of arrangements, for example :-

 Number of letters All letters different 1 letter repeated twice 2 letters repeated twice 3 letters repeated twice 4 24* 12* 6* - 5 120* 60* 30* - 6 720* 360 180 90 7 5040 2520 1260 630

* shown in detail in text; the remainder direct from the universal formula.

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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