There are five fruits - Banana, Apple, Orange, Pear and Mango - which are to be served among six boys. If no fruit can be served to more than one boy, and the fruits are served randomly, find the probability that all the five fruits were served to exactly three boys.PLEASE EXPLAIN THE APPROACH..this is an AIMCAT QUESTION.

1) 5/8
2) 25/54
3) 55/108
4) 125/324

• fruits can be distributed in 7 different ways..
  1. 113000 = 5C3*2C1*6! /(3!*2!) = 1200 [ 3! is for 3 boys who got Zero Fruits and 2! is for 2 boys who got 1 fruit each]
  
  2. 111110 = 5C1*4C1*3C1*2C1*1*6!/5! = 5!*6!/5! = 720
  
  3. 121100 = 5C2*3C1*2C1*6!/(3!*2!) = 3600
  
  4. 122000 = 5C2*3C2*1*6!/(3!*2!)= 1800
  
  5. 140000 = 5C4*1*6!/4! = 150
  
  6. 230000 = 5C3*2C2*6!/4! = 300
  
  7. 500000 = 5C5*6!/5! = 6
  
  Total= 1200+720+3600+1800+150+300+6 = 7776..
  Favorable cases are = Case no-1 and Case no-4 where Fruits are distributed among 3 boys..
  i.e. total favorable cases = 1200+1800 =3000
  
  Probability = 3000/7776 = 125/324 ............... ANSWER.
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