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?

A) 48 B) 64 C) 63 D) 45

Answer 1

Answer:

The box contains 2 White, 3 Black and 4 Red balls. To find - No. of ways of drawing 3 balls out of the box such that at least one black ball is included.

(Method 1) Let’s see the problem in this way

“At least one black ball” = “exactly 1 black ball and 2 non-black (White and Red) balls” or “exactly 2 black balls and 1 non-black (White and Red) balls” or “exactly 3 black balls and 0 non-black (White and Red) balls”,

Required no. of ways = No. of ways of selecting 1 black ball and 2 non-black balls (White and Red) + No. of ways of selecting 2 black balls and 1 non-black balls (White and Red) + No. of ways of selecting 3 black balls and 0 non-black balls (White and Red).

Hence the answer should be

= 3C1 ∗ 6C2 +3C2 ∗ 6C1 +3C3 ∗ 6C0

= 45 + 18 + 1 = 64

(Method 2) Let’s see the problem in this way

Required number of ways = Number of ways of selecting 3 balls out of given 9 balls (without any restriction) - Number of ways of selecting 3 balls when no black ball is to be selected.

Hence the answer should be

= 9C3 −6C3 = 84 - 20 = 64

please mark as brainliest.

Answer 2

B) 64

We 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 + 3 x 2 x 6 + 1

2 x 1 2 x 1

= (45 + 18 + 1)

= 64.

I think this answer is correct.