Sherlock homes and Dr Watson have to travel from Rajiv Gandhi Chowk to Indira Gandhi International airport via metro.they have enough coins of 1,5,10,25 paise. Sherlock agrees to pay for Dr Watson only if he tells all the possible combinations of coins that can be used to pay for the ticket.asked in Elitmus on 3rd August

Q.how many combinations are possible,if the fare is 50 paise.
a.52 b.49 c.45 d.44

• Let the number of 1, 5, 10 and 50 paise coins required are x, y, z and w respectively.
  
  So when paid total amount is: x + 5y + 10z + 25w, which is equal to the fare paid.
  
  So, x + 5y + 10z + 25w = 50
  
  The number of non negative integral solution of this equation will be equal to the number of ways the fare can be paid.
  
  
  
  x + 5y + 10z + 25w = 50
  
  In this equation w can be either 0, 1 or 2
  
  
  
  Case 1: When w is 2
  
  No other coins are required to pay
  
  So there is only one way to pay: (0, 0, 0, 2)
  
  Total no of ways : 1
  
  
  Case 2: When w is 1
  
  x + 5y + 10z = 25
  
  In this case z can have only three values possible: 0, 1 or 2
  
  Case I: When z is 0
  
  x + 5y = 25
  
  possible values are: (25, 0, 0, 1), (20, 1, 0, 1), (15, 2, 0, 1), (10, 3, 0, 1), (5, 4, 0, 1), (0, 5, 0, 1)
  
  Case II: When z is 1
  
  x + 5y = 15
  
  possible values are: (15, 0, 1, 1), (10, 1, 1, 1), (5, 2, 1, 1), (0, 3, 1, 1)
  
  Case III: When z is 2
  
  x + 5y = 5
  
  possible values are: (5, 0, 2, 1), (0, 1, 2, 1)
  
  Total no of ways : 12
  
  
  Case 3: When w is 0
  
  x + 5y + 10z = 50
  
  In this case z can have only three values possible: 0, 1, 2, 3, 4 or 5
  
  Case I: When z is 0
  
  x + 5y = 50
  
  possible values are: (50, 0, 0, 0), (45, 1, 0, 0), (40, 2, 0, 0), (35, 3, 0, 0), (30, 4, 0, 0), (25, 5, 0, 0), (20, 6, 0, 0), (15, 7, 0, 0), (10, 8, 0, 0), (5, 9, 0, 0), (0, 10, 0, 0)
  
  Case II: When z is 1
  
  x + 5y = 40
  
  possible values are: (40, 0, 1, 0), (35, 1, 1, 0), (30, 2, 1, 0), (25, 3, 1, 0), (20, 4, 1, 0), (15, 5, 1, 0), (10, 6, 1, 0), (5, 7, 1, 0), (0, 8, 1, 0)
  
  Case III: When z is 2
  
  x + 5y = 30
  
  possible values are: (30, 0, 2, 0), (25, 1, 2, 0), (20, 2, 2, 0), (15, 3, 2, 0), (10, 4, 2, 0), (5, 5, 2, 0), (0, 6, 2, 0)
  
  Case IV: When z is 3
  
  x + 5y = 20
  
  possible values are: (20, 0, 3, 0), (15, 1, 3, 0), (10, 2, 3, 0), (5, 3, 3, 0), (0, 4, 3, 0)
  
  Case V: When z is 4
  
  x + 5y = 10
  
  possible values are: (10, 0, 4, 0), (5, 1, 4, 0), (0, 2, 4, 0)
  
  Case VI: When z is 5
  
  x + 5y = 0
  
  possible values are: (0, 0, 5, 0)
  
  Total no of ways : 36
  
  
  So total no of ways = 1 + 12 + 36 = 49
• 10 years agoHelpfull: Yes(5) No(0)
• 49 is answr
• 10 years agoHelpfull: Yes(2) No(0)
• 49 as a+5b+10c+25d=50 ;; there are 49 combinations
• 10 years agoHelpfull: Yes(2) No(0)
• set on
  p+5p+10p+25p=50
  take combination of
  take 1p and rest will zero
  simlrly take 5p
  simlry take 10p
  simlrly take 125p
  again make comniation of 2 diffrnt
  agin make combination of 3 diffrnt
  last combination of 4
  then u will get total combination =52
• 10 years agoHelpfull: Yes(1) No(0)
• @aparna bajpai
  how is it 49? can u explain?
• 10 years agoHelpfull: Yes(0) No(1)