A and B have to travel from place x to place y.they have enough coins 1,5,10,25 paise.
A agrees to pay for B,only if he tells all the possible combinations of coins that can be used to pay for the ticket.

a.how many combinations are possible,if the fare is 50 paise?
b.how many combinations are possible ,if they get an discount of 10%,i.e., the fare is 45paise?

• their are 13 diff ways to get 25 like
  25
  10 10 5
  10 10 1 1 1 1 1
  10 5 5 5
  ......
  now by chosing this combination two time you can get 50
  here we can make solution like first with itself & other 12 (so 13) for the sencond it is the same so it will be (12) till 13th pair make combination with itself.....!
  so total count is 13+12+11+...1=91
• 12 years agoHelpfull: Yes(4) No(2)
• can you provide mathematical solution?
• 12 years agoHelpfull: Yes(3) No(0)
• a.how many combinations are possible,if the fare is 50 paise?
  
  48 combinations are possible.
• 12 years agoHelpfull: Yes(1) No(2)
• b.how many combinations are possible ,if they get an discount of 10%,i.e., the fare is 45paise?
  
  39 combinations are possible.
• 12 years agoHelpfull: Yes(1) No(2)
• there will be 13 combinations for 25..
  so first combination will pair up with all the 13..and the preceding combination will combine with next 12 to avoid repetition...and so on so it will be continued so final combination will be
  13+12+11+10+9+8+7+6+5+4+3+2+1=91
• 12 years agoHelpfull: Yes(0) No(1)
• a)take power set of {1,5,10,25}
  by doing for diff sets
  i.e as single elements 5 combinations(i.e 50 1's,10 5's....)
  (1,5)-no of 5's 1,2,3,4,5,6,7,8,9 totally 9 combinations
  (1,10)-no of 10's 1,2,3,4 -4 combinations
  |||ly doing v get 42 combinations...
• 12 years agoHelpfull: Yes(0) No(1)