Permutations question? (1 Viewer)

mathsfreak1010

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so im having a bit of trouble with this question:

A piece of art receives a mark out of 100 for each of the categories design, technique and originality. In how many ways is it possible to score mark 200/300


I dont even know how to start????
thanks in advance :D
 

braintic

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so im having a bit of trouble with this question:

A piece of art receives a mark out of 100 for each of the categories design, technique and originality. In how many ways is it possible to score mark 200/300


I dont even know how to start????
thanks in advance :D
Let A = mark for design, B = mark for technique, C = mark for originality

So we are solving the equation A+B+C=200, with the restriction that 0 <= A,B,C <= 100 and A,B,C are integers.

Consider A=0: there is only ONE possibility (B=100, C=100)

Consider A=1: there are TWO possibilities (B=99, C=100 or B=100, C=99)

Consider A=2: there are THREE possibilities (B=98, C=100 or B=99, C=99 or B=100, C=98)

.
.
.

Consider A=100: there are 101 possibilities (from B=0, C=100 through to B=100, C=0)


So the total number of possibilities is the sum of the integers 1 through 101.
The sum of AP formula gives 5151.







EDIT: Neater solution

Rather than distributing the 200 marks that are awarded, instead distribute the 100 marks that are NOT awarded.
That is, start by giving the artist 100/100 for each category, then figure out ways of deducting 100 marks.

Represent the 100 lost marks by a series of 100 dots, with two dividers separating the marks into the three categories.
To make the diagram manageable, first consider dividing 10 marks into 3 categories. Two possible scenarios are:

...|.....|.. (representing 3 marks for category A, 5 for B, 2 for C)
...||....... (3 marks for category A, zero for B, 7 for C)

So the number of possibilities reduces to the number of ways of forming a 'word' using 10 identical dots and 2 identical dividers.

This is (12!)/(10!.2!)

In the real question, the calculation is (102!)/(100!.2!) = 5151


Note that this is the ideal method when the total mark is 200 or more.
When the total mark is 100 or less, you would use the same method but on the AWARDED marks instead of the deducted marks.
However, when the total mark is between 100 and 200, you have a problem. This is because if you try to place the dividers anywhere without restriction, you encounter the possibility of awarding or deducting MORE than 100 marks for a particular category. Then you would have to take cases ... not many for 101 or 199, but LOTS for 150.
 
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