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Making Change for a Dollar (and other number partitioning problems)

dollar notes in denominations of 1, 2 and 5. How many ways are there to form exactly $100 using just multiples of these notes?

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marked as duplicate by Marc van Leeuwen, draks ..., Micah, rschwieb, froggie Nov 27 '12 at 14:43

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up vote 2 down vote accepted


$0\le x\le100\implies 0\le 100-2y-5z\le100\implies 0\le 2y+5z\le 100$ $\implies 0\le y\le \lfloor \frac{100-5z}2\rfloor$

As $0\le 5z\le 100, 0\le z\le 20 $

If $z$ is odd $=2a+1, 0\le y\le \lfloor \frac{100-5(2a+1)}2\rfloor=47-5a, 47-5a+1=48-5a$ values of $z$

So, $0\le 2a+1\le 20\implies 0\le a\le9 $

If $z$ is even,$=2b, 0\le y\le \lfloor \frac{100-5(2b)}2\rfloor=50-5b, 50-5b+1=51-5b$ values of $z$

So, $0\le 2b\le 20\implies 0\le b\le10 $

So, the number of combinations =$\sum_{ 0\le a\le9}(48-5a)+\sum_{0\le b\le10}(51-5b)=48\cdot10+51\cdot 11-5\sum_{ 0\le a\le9}a-5\sum_{0\le b\le10}b=541$

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THANKYOU for your help – chndn Nov 27 '12 at 11:39

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