Answer with Explanation:
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
= ( 3 x 6 x 5 / 2 x 1) + ( 3 x 2 /2 x 1 x 6 ) + 1
= (45 + 18 + 1)
= 64.
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
= ( 3 x 6 x 5 / 2 x 1) + ( 3 x 2 /2 x 1 x 6 ) + 1
= (45 + 18 + 1)
= 64.