Fourteen numbered balls (i.e. 1, 2, 3, ..., 14) are divided in three groups randomly. Find the probability that the sum of numbers on a balls, in each group, is odd.

Explanation

Each group should have odd numbered balls.
Case I: Two groups have three odd numbered balls and third has only one odd numbered ball.
Number of such cases = `(7!)/((3!)^(2)xx1!xx2!)xx3^(7)` (as even number of balls can go in any group)
Case II: Two groups have one odd numbered ball each and the third group has five odd numbered balls.
Number of such cases = `(7!)/((1!)^(2)xx5! xx 2!)xx3^(7)`
Total number of cases = Number of ways in which three non-empty group can be formed
`=(3^(14)-.^(3)C_(1)2^(14) + .^(3)C_(2))/(3!)`
`therefore` Required probability = `((7!)/(3!)^(2)+(7!)/(5! xx 2!))/(3^(14) - .^(3)C_(1)2^(14)+.^(3)C_(2))xx3!xx3^(7)`