3) I will now use different names with different numbers of letters and find out how many arrangements there are. I will then try to find a pattern.
MO:
RIA:
LUCY:
CASEY:
As you can see I haven’t written out all the arrangements for CASEY because this is not necessary. I found all the possible arrangements beginning with C and have multiplied it by the number of letters in the name which is 5.
So, 5 ÷ 24=120. This means there are 120 different arrangements in the name CASEY.
I think I have discovered a pattern:
If you look at MO, the number of arrangements is 2. For RIA the number of arrangements are 6 and for LUCY, 24. If you multiply the number of letters by the number of arrangements previously, then you will get the number of arrangements for that number of letter. Sounds confusing, here are a couple of examples.
No. of letters × no. of arrangements previously = no. of arrangements
RIA: 3 × 2 = 6
LUCY: 4 × 6 = 24
CASEY: 5 × 24 = 120
I can now use this equation to find out arrangements of other words with more letters.
To make this easier you could say:
1 × 2= 2 = 2!
1 × 2 × 3= 6 = 3!
1 × 2 × 3 × 4= 24= 4!
1 × 2 × 3 × 4 × 5= 120 = 5!
1 × 2 × 3 × 4 × 5 × 6= 720 = 6!
1 × 2 × 3 × 4 × 5 × 6 × 7= 5040 = 7!
1 × 2 × 3 × 4 × 5 × 6 × 7 × 8= 40320 = 8!
I will now see what happens when you have a name or word that has a letter in it that occurs twice or more and compare it to the same letters of a word where none of the letters repeat.
4 letters:
5 letters:
6 letters:
Again, I took the first letter from the words and tried to find as many arrangements and multiplied it by the number of letters in the words. I think I have found the equation.
η! = (no. of letters in factorial form)
χ! = (no. of letters repeating in factorial form)
Or to look at it in an easier way…
NAADUE: 720 ÷ 2 = 360
CAAAA: 120 ÷ 24 = 5
EMMA: 24 ÷ 2 = 12
I will now try to find out how many different arrangements there are with two sets of letters that occur more than once. E.g. ABBA
I’ll start off with ABBA because I like their music:
I will do AAABB next:
To predict how many different arrangements there was going to be, I tried to find out as many different arrangements beginning with B because I would have to find fewer arrangements as with A. I found 4. I divided this by two because there were two B’s in the word. So, 4 ÷ 2 = 2. I then multiplied this by 5 because there were 5 letters in the word. The answer was 10.
I think I have already worked out the equation for these sorts of arrangements.
ABBA: There are 4 letters in the word so I at first thought it might be 4! But this cannot be because there are two sets of different letters which occur more than once. I thought that there must be fewer arrangements because of this. So I thought:
ABBA = 4! ÷ (2! + 2!) = 6. This equation worked so I tried:
AAABB = 5! ÷ (3! + 2!) = 15. But this wasn’t right as there are only 10 arrangements.
At last I tried: 5! ÷ (3! × 2!)= 10. This worked. I tried it again with ABBA and it worked. Now I have found the equation:
No. of letters in factorial ÷ (No. of times a letter repeats × No. of times a letter repeats) = no. of arrangements
At the beginning of the course work I said that I wanted to be able to work out how many different arrangements there are in XXXYYVVVV by the end of the coursework. I can now do that: Arrangements =
η! 9! 362880 362880
(χ! × γ! × ν!) (3! × 2! × 4!) (6 × 2 × 24) 288
η! = 9! = no. of letters in the word
χ! = 3! = no. of x in the word
γ! = 2! = no. of Y’s in the word
ν! = 4! = no. of V’s in the word
Hope you enjoyed this thoroughly interesting piece of course work and I look forward to a good grade (PLEASE?) Bye!
By RIA Hylton 10A
