Emma’s Dilemma maths coursework - page 6
Keywords: Emma’s Dilemma maths coursework GCSE
By slashwk on 23/11/2006 17:25:42
Level: GCSE Key Stage 4 (Years 10-11)
Page Number: 6 of 6 pages: 1 2 3 4 5 6
to the right will produce twice as many repeated permutations. Using ANA as a starting example:
1. ANA
2. AAN
3. NAA
Therefore when an extra ‘N’ is added:
1. NANA
2. ANNA
3. ANNA
4. ANAN
5. NAAN
6. ANAN
7. AANN
8. AANN
9. NNAA
10. NNAA
11. NANA
12. NAAN
As you can see there are only six different combinations possible due to the repeated combinations caused by the repeated letters. As explained before repeated letters lessen the total amount of combinations as ‘CD’ can be written as ‘DC’ as well as ‘CD’, whereas ‘CC’ can only be written as ‘CC’.
Just to test my new formula we can apply it to a complex situation. Let’s take this one for example:
xxxxxxxxxxyyyyyzzzzzzzz (10 x’s, 5 y’s, 7 z’s)
C = n!
r!r2!r3!...
Our equation can be filled in like so:
C = 22!
10!5!7!
C = 1124000727777607680000
3628800 x 120 x 5040
C = 512143632
Our equation tells us that there are 512143632 different combinations with xxxxxxxxxxyyyyyzzzzzzz!
KEY
C The number of possible combinations
n The number of letters in that particular word or expression
r The amount of repetitions of any repeated letter (only applies to one letter, other letters must have separate ‘r’ value)
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