How many different combinations of quarters, nickels, and dimes can be used to make $0.55?
Quater =25Dime =10 Nickel =5In $0.55 max. 2 Quaters, max. 5 Dimes and max. 11 Nickels (2∑i=0x(25∗i))×(5∑i=0x(10∗i))×(11∑i=0x(5∗i))=(1+x25+x50)×(1+x10+x20+x30+x40+x50)×(1+x5+x10+x15+x20+x25+x30+x35+x40+x45+x50+x55)=(x155)+(x150)+2∗x145+2∗x140+3∗x135+4∗x130+5∗x125+6∗x120+7∗x115+8∗x110+10∗x105+11∗x100+11∗x95+12∗x90+12∗x85+13∗x80+13∗x75+12∗x70+12∗x65+11∗x60+11∗x55+10∗x50+8∗x45+7∗x40+6∗x35+5∗x30+4∗x25+3∗x20+2∗x15+2∗x10+(x5)+1
The coefficient from x55 is 11. So there are 11 possibilities.
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 ???
How many different combinations of quarters, nickels, and dimes can be used to make $0.55 ?
1.$0.55=3Dimes+1Quater2.$0.55=1Nickel+2Quaters3.$0.55=1Nickel+5Dimes4.$0.55=2Nickels+2Dimes+1Quater5.$0.55=3Nickels+4Dimes6.$0.55=4Nickels+1Dime+1Quater7.$0.55=5Nickels+3Dimes8.$0.55=6Nickels+1Quater9.$0.55=7Nickels+2Dimes10.$0.55=9Nickels+1Dime11.$0.55=11Nickels
Quater =25Dime =10 Nickel =5In $0.55 max. 2 Quaters, max. 5 Dimes and max. 11 Nickels (2∑i=0x(25∗i))×(5∑i=0x(10∗i))×(11∑i=0x(5∗i))=(1+x25+x50)×(1+x10+x20+x30+x40+x50)×(1+x5+x10+x15+x20+x25+x30+x35+x40+x45+x50+x55)=(x155)+(x150)+2∗x145+2∗x140+3∗x135+4∗x130+5∗x125+6∗x120+7∗x115+8∗x110+10∗x105+11∗x100+11∗x95+12∗x90+12∗x85+13∗x80+13∗x75+12∗x70+12∗x65+11∗x60+11∗x55+10∗x50+8∗x45+7∗x40+6∗x35+5∗x30+4∗x25+3∗x20+2∗x15+2∗x10+(x5)+1
The coefficient from x55 is 11. So there are 11 possibilities.