Question Number 117192 by Lordose last updated on 10/Oct/20
$$\mathrm{Assuming}\:\mathrm{you}\:\mathrm{have}\:\mathrm{enough}\:\mathrm{coins}\: \\ $$$$\mathrm{of}\:\mathrm{1},\mathrm{5},\mathrm{10},\mathrm{25},\mathrm{and}\:\mathrm{50cents}.\:\mathrm{In}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{you}\:\mathrm{make}\:\mathrm{a}\:\mathrm{change}\:\mathrm{for}\:\mathrm{1dollar}. \\ $$
Answered by mr W last updated on 10/Oct/20
$${a}+\mathrm{5}{b}+\mathrm{10}{c}+\mathrm{25}{d}+\mathrm{50}{e}=\mathrm{100} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)× \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} +{x}^{\mathrm{15}} +…\right)× \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{10}} +{x}^{\mathrm{20}} +{x}^{\mathrm{30}} +…\right)× \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{25}} +{x}^{\mathrm{50}} +{x}^{\mathrm{75}} +…\right)× \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{50}} +{x}^{\mathrm{100}} +{x}^{\mathrm{150}} +…\right) \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{x}^{\mathrm{5}} \right)\left(\mathrm{1}−{x}^{\mathrm{10}} \right)\left(\mathrm{1}−{x}^{\mathrm{25}} \right)\left(\mathrm{1}−{x}^{\mathrm{50}} \right)} \\ $$$${coefficient}\:{of}\:{term}\:{x}^{\mathrm{100}} \:{is}\:\mathrm{292},\:{i}.{e}. \\ $$$${we}\:{have}\:\mathrm{292}\:{ways}\:{to}\:{change}\:\mathrm{1}\:{dollar}. \\ $$
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
$${i}\:{used}\:{generating}\:{function}\:{method} \\ $$$${to}\:{find}\:{the}\:{number}\:{of}\:{non}−{negative} \\ $$$${integer}\:{solutions}\:{of}\:{equation} \\ $$$${a}+\mathrm{5}{b}+\mathrm{10}{c}+\mathrm{25}{d}+\mathrm{50}{e}=\mathrm{100} \\ $$
Commented by Lordose last updated on 10/Oct/20
Thanks sir ��