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Maths question help - permutations?


RyanJ

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Hi there!

 

I have this question and despite my best efforts I have failed to answer it.

 

Given an unlimited supply of 50p, £1 and £2 coins, in how many different ways is it possible to make a sum of £100.

a) 1,326

b) 2,500

c) 5,050

d) 10,000

 

I've guessed that is it 5,050 I have no idea how to solve this - any help appreciated!

 

Cheers,

 

Ryan Jones

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Let $2 and $1 be the determining factor and let $0.50 to mark up the sum to $100.

 

When $2 x 50, $1 x 0 and $ 0.50 x 0~~~~~this is one way

 

When $2 x 49, $1 can be x1 or x2 or x0 and let $0.50 to mark up to the sum

Then this will be 3 way. $ 1 x0 would mean than no one dollar is used, only two dollar and 50p will be used.

 

When $2 x 48, $1 can be x1 or x2 or x3 or x4 or x0. Therefore, there are 5 ways

..............................

..............................

when $2 x 1, $1 can be x0 or x1 or .... or x 98, a total of 99 ways

when $2 x 0, $1 can be x0 or x1 or .... or x 98 or x 99 or x 100, a total of 101 ways

 

As you can see , this is an arithmatic progression. To find possible way to make a sum of $ 100, apply arithmatic summation.

 

It will be 1+3+5+...+101=2601

 

I don't get answer from the option. Maybe there is some mistake in my calculation? Please kindly point out.

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I think 2601 is actually the correct answer:

 

There will be 51 possible values of £2, each time leaving a left over of 0,2,4,6...98,100.

 

Each of these values will be made up of a some £1 and an even number of 50p coins. For each value there is value+1 ways of making it up. So we have to solve:

 

1 + 3 + 5 + .... + 99 + 101

 

The arithmetic series can be totalled with this formula:

 

[math]S_n = \tfrac{n}{2}(a + l)[/math]

 

Where n is the number of values, a is the first value and l is the last. So:

 

[math]S_n = \frac{51 * (1+101)}{2}[/math]

 

Which equals 2601.

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I've got a possible answer to why the question may be wrong. If you use the summation technique posted by builgate you must have the answer to be:

 

[math]\sum_{n=0}^{50} (2n+1) = 51^2 = 2601[/math].

 

It's quite probable that someone made a numerical mistake, perhaps summing (2n-1) to give an answer of 502 = 2500 by mistake. I've looked at both the methods and can't see a problem, so... I don't know :)

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Thanks for the help guys!

 

This was from a senior maths callange - how are you suposed to work this out in your hear in 20 seconds is what I'd like to know!

 

Oh and sorry 2,500 was supposed to be 2,601 :)

 

Cheers,

 

Ryan Jones

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