Q. Sherlock and Waston have to travel from Rajiv Gandhi Chowk to airport in metro. they have enough coins 1,5,10,25 paise. Sherlock agree to pay for Waston only if he tells all the possible combination of coins that can used to pay for the ticket.

1. how many combination are possible if the fare is 50 paise.
2. how many combination are possible if the discount on fair is 10% ie fare is 45paise

• For 1st case: Answer is 49.
  25|10|5|1|Total
  ----------------
  0|0|0|50--1 way
  0|0|1-10|45-0--10 ways
  0|1|0-8|40-0--9 ways
  0|2|0-6|30-0--7 ways
  0|3|0-4|20-0--5 ways
  0|4|0-2|10-0--3 ways
  0|5|0|0--1 way
  1|0|0-5|25-0--6 ways
  1|1|0-3|15-0--4 ways
  1|2|0-1|5-0--2 ways
  2|0|0|0--1 way
  
  so, the total no of ways is- 49 ways.
• 11 years agoHelpfull: Yes(3) No(0)
• I think for 1 its 30 possible cases as follow
  
  cases: 25 including (13)
  
  25+25 (25+1*25)
  25+10+10+5(25+10+10+1*5)(25+10+5+1*10)(1*25+10+10+5)
  25+10+5+5+5(again 3 case)
  25+5+5+5+5+5(2 case as it has two coins two replace by 1)
  
  cases: excluding 25 only (14)
  
  10+10+10+10+10 (10+10+10+10+1*10)
  10+10+10+10+5+5 (10+10+10+10+5+1*5)(10+10+10+1*10+5+5)
  10+10+10+5+5+5+5 (again 2 case)
  10+10+5+5+5+5+5+5 (again 2 case)
  10+5+5+5+5+5+5+5+5 (again 2 case)
  
  cases: excluding 25,10 (2)
  
  5+5+5+5+5+5+5+5+5+5 (only 1 case)
  
  cases : only 1rs coin (1)
  
  1*50
  
• 11 years agoHelpfull: Yes(2) No(3)
• in no 1 question ... total possible cases are 49.
  in no.2 question... total possible cases are 26.
  
• 10 years agoHelpfull: Yes(1) No(1)
• @ bhaskar pal couldnt understand
• 11 years agoHelpfull: Yes(0) No(0)
• pls tell clearly
• 11 years agoHelpfull: Yes(0) No(0)
• 25+25
  25+10+10+5
  25+10+5+5+5
  25+5+5+5+5+5
  25+5+5+5+5+1*5
  25+5+5+5+1*5+1*5
  .
  .
  .
  .
  10+10+10+10+10
  10+10+10+10+5+5
  10+10+10+5+5+5+5
  .
  .
  .
  .
  10+1*5+1*5........
  .
  .
  .
  5*10
  5*9+1*5
  5*8+1*5+1*5
  ;.
  .
  .
  .
  .
  1*50
  answer is 33.
• 10 years agoHelpfull: Yes(0) No(2)
• case 25(fixed)::
  +25 1
  cases number of cases
  +10+10+5(/or)1*5 2
  +10+5+5+5(/)1*5 2
  +10+5+1*5+5/1*5 2
  +10+1*5+(5+5(/)1*5)(/or)(1*5+5/1*5) 4
  
  and so on will give 50 possibilities and then if we vary 25(first half) again then it will give 50*50=2500.
• 10 years agoHelpfull: Yes(0) No(1)
• for case 1
  answr should b 53c3
  becoz we tak it as 1x+5y+10z+25w=50
  hence total numbr of solutions are 53c3
• 10 years agoHelpfull: Yes(0) No(0)