Emma's Dilemma

How many combinations would there be if three letters were used?

For example, Ted.

Ted: 3 letters, none the same.

. Ted

2. Tde

3. Dte

4. Edt

5. Det

6. Etd

For Ted, there are 6 possible arrangements. There are three letters as such the combination will always have three staring points, so each letter starts the word twice. 3(the number of letters)x2(the number of times a letter starts the name)=6.

What if there were 4 letters with 2 pairs?

I shall try the name Alla.

Arrangements for alla: 4 letters, two pairs

.alla

2.aall

3.llaa

4.alal

5.lala

6.laal

There are 6 arrangements for Alla. From 2 pairs of letters to all different letters in a 4-letter word I have found a pattern of 6, 12 and 24. As it is often far easier to see what is happening with more difficult problems, I shall investigate a name with 5 letters.

I have found that there were 24 arrangements for a 4 letter word with all different letters, I have also found that there were 6 possible ways of starting a 4 letter word with each letter. I predict that there will be 120 arrangements for a 5 letter word with all different letters. The reasons I predict 120 is because with four letters (all different) there were 24 combinations, and with 5 letters all that is happening is that an extra point to start form is added. Eg 24 combinations a 4 letter word, with an extra in front of it. So 120=24

5

Previously in the 4-letter word, 24 divided by 4 equals 6, the number of possibilities there were for each letter.

To make things simpler and east to under stand I shall use the name Xlucy.

.xlucy 2.xucyl 3.xcylu 4.xycul

5.xluyc 6.xucly 7.xcyul 8.xyclu

9.xlcuy 10.xulcy 11.xculy 12.xyulc

3.xcyu 14.xulyc 15.xcuyl 16.xyucl

7.xlyuc 18.xuycl 19.xclyu 20.xylcu

21.xlycu 22.xuylc 23.xcluy 24.xyluc

I have done the first 24 in this sequence, as it makes it easy to see that to obtain the other combinations all that is needed is to replace the first letter with one of the others, so I will not list all 120 possibilities.

I can see that the numbers of possibilities for different arrangements are going to increase as more letters are used, as long as the letters are different. I have come up with a formula that can be used to deduce the number of possibilities with out having to list them. For lucy's name, with four letters 4x3x2x1. For Ted 3x2x1, and for Xlucy 5x4x3x2x1.

Luckily scientific calculators have a button for doing this. It is called factorising. The button Is marked as an exclamation mark. If for example, I type in 5! On the calculator it will give the answer of 120, which makes sense as 120 divided by 5 equals 24, the number of possibilities for a 4 letter word.

So as a general formula for names with x number of letters all different I have come up with a formula. With Lucy's name; 1x2x3x4 = 24. With qlucy; 1x2x3x4x5 = 120. However this is expressed as factorial. There is a button on most scientific calculators with have embedded this factorial button feature generally sowing as an exclamation mark. All it does is save the time of having to put in to the calculator 1x2x3x4x5x6x7..... Ect. You just put in the number and press factorial and it will do 1x2x3... until it get to the number you put in. If I key in (lets say the number of letters all different) factorial 6 I get it gives me 720, with makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word and it continues to fall in that pattern.

Factorising works for name's with no repeated letters, but the name Emma has four letters, two of which are the same. 4 factorial is 24, but I could only find 12 possibilities, this means that factorising will not work alone. There for if there are repeated letters, you answer must be divided by the number of repeated letters, eg Emma, 4!= 24 24=12 the number of

2

combinations .

It is easier to understand in terms of x And y. I will start with xxyy:

Arrangements for xxyy: 4 letters, 2 pairs

.xxyy 2.xyxy 3.xyyx

4.yxxy 5.yxyx 6.yyxx

There are 6 arrangements.

Arragements for xxxyy: 5 letters, a pair and 3 the same

.xxxyy 2.xxyxy 3.xxyyx 4.xyxyx 5.xyxxy

6.xyyxx 7.yyxxx 8.yxxxy 9.yxyxx 10.yxxyx

There are 10 different arrangements .

arrangement for xxxxy:

.xxxxy 2.xxxyx 3.xxyxx 4.xyxxx 5.yxxxx

There are 5 different arrangements for this instance.

If I use xxxyy as an example, there are 2y's and 3 x's in a total of 5 unknowns. The best formula I could come up with is the number of total letters factorial, divided by the number of x's, y's etcfactorised and multiplied.

eg

A five letter word like xxxyy; this has 3 x's and 2 y's. 1x2x3x4x5 = 120 / 12 = 10

x2x3 x 1x2