permutations and combinations

Question

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

Comments

3C₁ *6C₂ + 3C₂ * 6C₁  + 3C3 * 6C₀

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why are we considering balls as distinct??

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i did nt consider thm as different.i am using combination,not permutation

Answer 1

At least one black ball is to be included, So there can be three cases,

(1 black ball $ \& $ 2 other ball) + (2 black ball $ \& $ 1 other ball) + (3 black ball $ \& $ 0 other ball)

 $( 3C_1 * 6C_2 ) + ( 3C_2 * 6C_1 ) + ( 3C_3 * 6C_0 )$

Answer 2

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

2 white balls

3 black balls

4 red balls.

white   black   red (atleat one black ball)

0           3          0

0           2           1

1           2           0

1           1            1

3C3 + 2C2 * 4C1 + 2C1*3C2 +2C1 * 3C1 * 4C1=1 + 4 + 6 +24=35