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Question Number 117192 by Lordose last updated on 10/Oct/20

$$\mathrm{Assuming}\mathrm{you}\mathrm{have}\mathrm{enough}\mathrm{coins}$$$$\mathrm{of}1,5,10,25,\mathrm{and}50\mathrm{cents}.\mathrm{In}\mathrm{how}$$$$\mathrm{many}\mathrm{ways}\mathrm{can}\mathrm{you}\mathrm{make}\mathrm{a}\mathrm{change}\mathrm{for}1\mathrm{dollar}.$$

Answered by mr W last updated on 10/Oct/20

$$a+5b+10c+25d+50e=100$$$$(1+x+x^2+x^3+...)\times$$$$(1+x^5+x^{10}+x^{15}+...)\times$$$$(1+x^{10}+x^{20}+x^{30}+...)\times$$$$(1+x^{25}+x^{50}+x^{75}+...)\times$$$$(1+x^{50}+x^{100}+x^{150}+...)$$$$=\frac{1}{(1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})}$$$$coefficientofterm{x}^{100}is292,i.e.$$$$wehave292waystochange1dollar.$$

Commented by Lordose last updated on 10/Oct/20

Thanks sir

Commented by Lordose last updated on 10/Oct/20

$$\mathrm{can}\mathrm{you}\mathrm{explain}\mathrm{sir}?$$$$$$

Commented by Lordose last updated on 10/Oct/20

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Commented by mr W last updated on 10/Oct/20

$$iusedgeneratingfunctionmethod$$$$tofindthenumberofnon-negative$$$$integersolutionsofequation$$$$a+5b+10c+25d+50e=100$$

Commented by Lordose last updated on 10/Oct/20

Thanks sir ��

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