serlok homes and watson have to travel from rajiv chowk to airport by metro.they have enough amount of 1,5,25 paise coin. homes agrees to pay for watson only if he tells the possible combination of coin used to pay for the ticket.
1. how many combinations are possible if the fair is 50 paise??
2. how many combinations are possible if the get discount of 10%. i.e the fare is 45 paise??

• Let us assume that there are x,y and z number of coins of 1 paisa, 5 paisa and 25 paisa. then according to question we can write x+5y+25z=50.
  so possible value for z will be 0,1and2. when z=o i.e x+5y=50 hence possible pair for x and y will be (0,10),(5,9),(10,8)........(50,0) total is 11.
  when z=1 i.e x+5y=25 hence possible pair for x and y will be (0,5),(5,4),(10,3)........(25,0) total is 6.
  when z=2 i.e x+5y=0 hence possible pair for x and y will be only (0,0)total number is 1.
  1)hence total possible combination will be 11+6+1=18.
  and for 2)....according to question we can write x+5y+25z=45.
  so possible value for z will be 0 and1.
  when z=o i.e x+5y=45 hence possible pair for x and y will be 10
  and when z=1 i.e x+5y=20 hence possible pair for x and y will be 5.
  2)hence total possible combination will be 10+5=15.
  
• 10 years agoHelpfull: Yes(17) No(0)
• lets first consider the cases of 1,5 and 10. We do not consider the cases of 25, as of now
  If 10 coins = 5, 5 and 1 coins = 0, 1 possibility
  If 10 coins = 4, 5 coins = 0,1,2 3 possibilities
  If 10 coins = 3, 5 coins = 0,1,2,3,4 5 possibilities
  If 10 coins = 2, 5 coins = 0,1,2,3,4,5,6 7 possibilities
  If 10 coins = 1, 5 coins = 0 to 8 9 possibilities
  If 10 coins = 0, 5 coins = 0 to 10 11 possibilities
  
  36 possibilities
  
  Now the case 25 paise coins
  25 coins = 2 1 possibility
  25 coins = 1 now we consider the sub-possibilitites of this case
  
  10 coins = 2, 5 coins= 0,1 2 possibilities
  10 coins =1, 5 coins = 0,1,2,3 4 possibilities
  10 coins = 0, 5 coins = 0,1,2,3,4,5 6 possibilities
  
  Total 13 possibilities from this case
  
  Total possibilities = 36 +13 = 49 possibilities
• 11 years agoHelpfull: Yes(8) No(3)
• Now the case of fare of 45 paise
  lets first consider the cases of 1,5 and 10. We do not consider the cases of 25, as of now
  If 10 coins = 4, 5 coins = 0,1 2 possibility
  If 10 coins = 3, 5 coins = 0,1,2,3 4 possibilities
  If 10 coins = 2, 5 coins = 0,1,2,3,4,5 6 possibilities
  If 10 coins = 1, 5 coins = 0,1,2,3,4,5,6,7 8 possibilities
  If 10 coins = 0, 5 coins = 0 to 9 10 possibilities
  
  30 possibilities
  
  Now the case 25 paise coins
  25 coins = 1 now we consider the sub-possibilitites of this case
  
  10 coins = 2, 5 coins= 0 1 possibilities
  10 coins =1, 5 coins = 0,1,2 3 possibilities
  10 coins = 0, 5 coins = 0,1,2,3,4 5 possibilities
  
  Total 9 possibilities from this case
  
  Total possibilities = 30 +9 = 39 possibilities
• 11 years agoHelpfull: Yes(2) No(4)
• sorry they have enough coins of 1,5,10&25 paise
• 11 years agoHelpfull: Yes(1) No(2)
• (15)
  x+5y+25z=50
  at z=0, x+5y=50 then y can have 10 possible values.
  at z=1 x+5y=25 then y can have 5 possible values.
  at z=2 x+5y=0 then y can have 0 possible values.
  so total combinations will be 15.
• 10 years agoHelpfull: Yes(1) No(0)