Mathematics Coursework : The Phone Box Problem

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Mathematics Coursework : The Phone Box Problem

Introduction: A woman only has 10 and 20 pence coins to use in a phone box. I have to investigate the number of different ways she could use 10 and 20 pence coins in a phone box. I am also investigating a man who only has 10 and 50 pence coins. I will start off with the woman.
10 pence call
                                             10=10p =1 way
20 pence call
20=10p+10p
                                              20=20p =2 ways
30 pence call
30=10p+10p+10p
                                          30=20p+10p =3 ways
30=10p+20p
40 pence call
40=10p+10p+10p+10p
40=20p+10p+10p
                                      40=10p+20p+10p =5 ways
40=10p+10p+20p
40=20p+20p
50 pence call
50=10p+10p+10p+10p+10p
50=20p+10p+10p+10p
50=10p+20p+10p+10p
                                 50=10p+10p+20p+10p =8ways
50=10p+10p+10p+20p
50=20p+20p+10p
50=20p+10p+20p
50=10p+20p+20p

Pascal´s triangle theory
                                       This should help me to work out long and write out long equations like the ones I am doing

E.g. 50 5 10p =1
                           1 20p and 3 10p=4 =8 ways
2 20p and 1 10p=3

To get this I used the calculator button which has a big c you put the total number of units at the top then you put the number of 20 pence coins at the bottom. You use the biggest one at the bottom I use 20 pence because it is relevant. The way that Pascal made up this theory was he got a triangle and started with 1 at the top and started adding them up as he went down like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
He did this for the rest of the way down. Instead of a calculator Pascal would have used the number of units added up to see how far he had to go down and the number of units in the biggest value to see how far he would of gone across.
E.g.50
5 10p =1
1 20p and 3 10p=4 down and 1 across
2 20p and 1 10p=3 down and 2 across
   You then look at where the numbers would be on the triangle and add them up that would come to 1+4+3 which would equal 8.

Investigation
                      I am now going to use Pascal´s triangle in my investigation to shorten the length of my equations.
60 pence call
60= 6 10p =1
60=1 20p and 4 10p=5
60=2 20p and 2 20p=6
60=3 20p=1
This would equal 13 different combinations
70 pence call
70=7 10p=1
70=1 20p and 5 10p=6
70=2 20p and 3 10p=10
70=3 20p and 1 10p=4
This would equal 21 different combinations
80 pence call
80=8 10p=1
80=1 20p and 6 10p=7
80=2 20p and 4 10p=15
80=3 20p and 2 10p=10
80=4 20p=1
This would equal 34 different combinations
Results
            This is a table of my results and a prediction for this far into this investigation
Cost of phone call 10 pence 20 pence 30 pence 40 pence 50 pence 60 pence 70 pence 80 pence
number of combinations 1 2 3 5 8 13 21 34

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By these results I would say this was the Fibinachi series which means to work out the next one you have to add the two previous terms to work out the next term. You can right it as the equation:
Tn=(Tn-1)+(Tn-2)
Tn means term number

Prediction
               I would say using the Fibinachi series the next number for a 90 pence call would be 21+34 so it would be 55

90 pence call
90=9 10p=1
90=1 20p and 7 10p=8
90=2 20p and 5 10p=21
90=3 20p and 3 10p=20
90=4 20p and 1 10p=5
This would equal 55 different combinations
Conclusion
                 My prediction was right that a 90 pence call would ...

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