A BOY HAS INITIALLY 2 RUPESE.HE PLAYS A GAME
.HE GAINS 1 IF WINS AND LOSES 1 IF LOSES
.HE CAN LOSE MAXIMUM 3 GAMES.IN HOWMANY WAYS HE CAN HAVE A TOTAL OF 5 RUPEES.

• when he loses 3 games he has to win 6 games
  this can be achieved in 9!/(6!*3!)=84 ways
  when he loses 2 games he needs to win 5
  no of ways= 7!/(5!*2!)=21
  similarly losing 1 & winning 4 = 5!/4!=5
  winning all in 1 way
  so total ways= 84+21+5+1=111
• 12 years agoHelpfull: Yes(12) No(0)
• SOLUTION:
  9C6 + 7C5 + 5C4 +3c3 = 111
  To get total 5 at end, the possibilities are:
  winning 6 out of 9 games : 9C6 (since,we can't loose more than 3 games)
  winning 5 out of 7 games : 7C5
  winning 4 out of 5 games : 5C4
  winning 3 out of 3 games : 3C3 (winning all match)
  
  
• 12 years agoHelpfull: Yes(7) No(0)
• when he loses 3 games he has to win 6 games
  this can be achieved in 9!/(6!*3!)=84 ways
  when he loses 2 games he needs to win 5
  no of ways= 7!/(5!*2!)=21
  similarly losing 1 & winning 4 = 5!/4!=5
  winning all in 1 way
  so total ways= 84+21+5+111
• 12 years agoHelpfull: Yes(3) No(2)