+0

# How many different combinations of quarters, nickels, and dimes can be used to make $0.55? 0 667 3 How many different combinations of quarters, nickels, and dimes can be used to make$0.55?

Guest Apr 21, 2015

#3
+20038
+5

$$\small{\text{Quater  = 25 \qquad Dime  = 10 \qquad  Nickel = 5 }} \\ \small{\text{In \0.55 max. 2 Quaters, max. 5 Dimes and max. 11 Nickels }} \\\\ \left(\sum\limits_{i=0}^{2} x^{({25}*i}) \right) \times \left (\sum\limits_{i=0}^{5} x^{({10}*i)} \right) \times \left(\sum\limits_{i=0}^{11} x^{({5}*i)} \right) \\\\ \small{\text{=(1+x^{25}+x^{50})}} \times \small{\text{(1+x^{10}+x^{20}+x^{30}+x^{40}+x^{50})}} \\ \times \small{\text{(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}+x^{30}+x^{35}+x^{40}+x^{45}+x^{50}+x^{55})}}\\ \small{\text{ =(x^{155})+(x^{150})+2*x^{145}+2*x^{140}+3*x^{135}+4*x^{130} }} \\ \small{\text{+5*x^{125}+6*x^{120}+7*x^{115}+8*x^{110}+10*x^{105}+11*x^{100} }} \\ \small{\text{+11*x^{95}+12*x^{90}+12*x^{85}+13*x^{80}+13*x^{75}+12*x^{70} }} \\ \small{\text{+12*x^{65}+11*x^{60}+{11*x^{55}}+10*x^{50}+8*x^{45}+7*x^{40} }} \\ \small{\text{+6*x^{35}+5*x^{30}+4*x^{25}+3*x^{20}+2*x^{15}+2*x^{10}+(x^5)+1 }} \\$$

The coefficient from  $$\small{\text{{x^{55}}}}$$ is $$\small{\text{{ 11 }}}$$. So there are 11 possibilities.

heureka  Apr 21, 2015
#1
+90180
+5

Let's see.......

2 quarters  1 nickel

1 quarter  3 dimes

1 quarter 2 dimes  2 nickels

1 quarter  1 dime 4 nickels

1 quarter 6 nickels

5 dimes 1 nickel

4 dimes 3 nickels

3 dimes 5 nickels

2 dimes 7 nickels

1 dime 9 nickels

11 nickels

I think that's it......did I leave anything  out ???

CPhill  Apr 21, 2015
#2
+20038
+5

How many different combinations of quarters, nickels, and dimes can be used to make \$0.55 ?

$$\small{\text{  \begin{array}{rccrlclcl} \hline \\ 1. & \0.55 &=& & && 3\; Dimes &+& 1\; Quater \\ 2. & \0.55 &=& 1& Nickel && &+& 2\; Quaters \\ 3. & \0.55 &=& 1& Nickel &+& 5\; Dimes \\ 4. & \0.55 &=& 2& Nickels &+& 2\; Dimes &+& 1\; Quater \\ 5. & \0.55 &=& 3& Nickels &+& 4\; Dimes \\ 6. & \0.55 &=& 4& Nickels &+& 1\; Dime &+& 1\; Quater\\ 7. & \0.55 &=& 5& Nickels &+& 3\; Dimes \\ 8. & \0.55 &=& 6& Nickels && &+& 1\; Quater \\ 9. & \0.55 &=& 7& Nickels &+& 2\; Dimes \\ 10. & \0.55 &=& 9& Nickels &+& 1\; Dime \\ 11. & \0.55 &=& 11& Nickels \\ \\ \hline \end{array} }}$$

heureka  Apr 21, 2015
#3
+20038
+5

$$\small{\text{Quater  = 25 \qquad Dime  = 10 \qquad  Nickel = 5 }} \\ \small{\text{In \0.55 max. 2 Quaters, max. 5 Dimes and max. 11 Nickels }} \\\\ \left(\sum\limits_{i=0}^{2} x^{({25}*i}) \right) \times \left (\sum\limits_{i=0}^{5} x^{({10}*i)} \right) \times \left(\sum\limits_{i=0}^{11} x^{({5}*i)} \right) \\\\ \small{\text{=(1+x^{25}+x^{50})}} \times \small{\text{(1+x^{10}+x^{20}+x^{30}+x^{40}+x^{50})}} \\ \times \small{\text{(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}+x^{30}+x^{35}+x^{40}+x^{45}+x^{50}+x^{55})}}\\ \small{\text{ =(x^{155})+(x^{150})+2*x^{145}+2*x^{140}+3*x^{135}+4*x^{130} }} \\ \small{\text{+5*x^{125}+6*x^{120}+7*x^{115}+8*x^{110}+10*x^{105}+11*x^{100} }} \\ \small{\text{+11*x^{95}+12*x^{90}+12*x^{85}+13*x^{80}+13*x^{75}+12*x^{70} }} \\ \small{\text{+12*x^{65}+11*x^{60}+{11*x^{55}}+10*x^{50}+8*x^{45}+7*x^{40} }} \\ \small{\text{+6*x^{35}+5*x^{30}+4*x^{25}+3*x^{20}+2*x^{15}+2*x^{10}+(x^5)+1 }} \\$$

The coefficient from  $$\small{\text{{x^{55}}}}$$ is $$\small{\text{{ 11 }}}$$. So there are 11 possibilities.

heureka  Apr 21, 2015