Three red counters, four green counters and five blue counters are placed in a row in random order. Find the probability that no two blue counters are adjacent.

1) 7/99
2) 13/198
3) 5/99
4) 11/198

• total number of ways of arranging with no blue near =(8c5 *7! / (3!*4!) )
  total arrangement =(12!/(3!*4!*5!)
  
  so ans =(8c5 *7! / (3!*4!) ) / (12!/(3!*4!*5!)
• 11 years agoHelpfull: Yes(1) No(3)
• no of favourable outcome= 8c5
  total no of outcomes= 12c5
  probability= 8c5/12c5 = 7/99
• 10 years agoHelpfull: Yes(1) No(0)