THERE ARE SOME COINS IN THE BAG(1 PAISE,5 PAISE,10 PAISE,25 PAISE).A MAN HAS TO DISTRIBUTE COINS TO THE CHILDREN IN SUCH A WAY THAT EACH ONE GET 50 PAISE exact HAVING NO TWO GETTING THE SAME DENOMINATION OF COINS.THEN HOW MANY CHILDRENS ARE THERE?

• 8 children?
  I have approached it in this way:
  Taking single coin combo:
  1) 5*10 = 50
  2) 10*5 = 50
  3) 25*2 = 50
  ---------------------
  Taking (5,10) coin combo:
  4) 5*2 + 10*4 = 50
  5) 5*4 + 10*3 = 50
  6) 5*6 + 10*2 = 50
  7) 5*8 + 10*1 = 50
  ---------------------
  Taking (10,25) coin combo:
  -> Not possible to make 50
  (in any new combo)
  ----------------------
  Taking (5,25) coin combo:
  8 ) 5*5 + 25*1 = 50
  Ergo,8 children.
  P.S : I need verification for this
  approach and also for the answer..!!
• 11 years agoHelpfull: Yes(4) No(3)
• @ SHAZM
  cant there be comb as (5,10,25),(1,10,25),(1,5,25),(1,5,10)etc,,,,???
  
  
• 10 years agoHelpfull: Yes(1) No(0)
• no two getting same denomination means?
• 11 years agoHelpfull: Yes(0) No(0)
• Arrange the combinations of the coins in the decreasing order of denominations :
  
  
  Case 1: when there is 2 25pc coin - 1 ways
  
  
  Case 2: when there is 1 25pc coin and
  (1) 2 10pc coin - nos of 5pc coin will range from 1 to 0 i.e 2 combinations
  (2) 1 10pc coin - nos of 5pc coin will range from 3 to 0 i.e 4 combinations
  (3) 0 10pc coin - nos of 5pc coin will range from 5 to 0 i.e 6 combinations
  
  
  Case 3: when there is no 25 pc coins and
  (1) 5 10pc coin - 1 ways
  (2) 4 10pc coin - nos of 5pc coin will range from 2 to 0 i.e 3 combinations
  (3) 3 10pc coin - nos of 5pc coin will range from 4 to 0 i.e 5 combinations
  (4) 2 10pc coin - nos of 5pc coin will range from 6 to 0 i.e 7 combinations
  (5) 1 10pc coin - nos of 5pc coin will range from 8 to 0 i.e 9 combinations
  (6) 0 10pc coin - nos of 5pc coin will range from 10 to 0 i.e 11 combinations
  
  
  
  So total 49 different combinations and hence 49 childrens
• 7 years agoHelpfull: Yes(0) No(0)