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

A. 16
B. 32
C. 64
D. 96
E. 128


Solution:

Without any restrictions, the number of ways to choose 3 balls from 9 is 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84.

Let’s calculate the number of ways we don’t select a black ball. Since there are 6 non-black balls, we can choose any 3 of them in 6C3 = (6 x 5 x 4) / (3 x 2) = 120/6 = 20 ways.

In summary, there are 84 different selections of 3 balls (if there are no restrictions), and 20 of them consist of selecting no black balls. Therefore, there must be 84 - 20 = 64 different selection of 3 balls with at least one black ball.

Answer: C
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A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

Total 9 balls are there ( 2W, 3B,4R)

Total no of ways to select 3 balls out of 9 = 9C3 = 9*8*7/3*2*1 = 84

Total no of ways to select 3 balls ,where there is no black ball selected i.e you should select from either white or red balls ( 2W + 4R) = 6C3= 20

Total no of ways to select 3 balls if atleast 1 black ball is included = Total no of ways to select 3 balls out of 9 - Total no of ways to select 3 balls ,where there is no black ball selected = 84 -20 = 64

Option C is the answer.

Thanks,
Clifin J Francis.
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Total no. ways to select atleast 1 black ball = total ways of selection of any 3 balls from 9 balls without any restriction - total ways of selection 3 balls from group of only red and white balls
=9C3-6C3
=84-20
=64
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Bunuel
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

A. 16
B. 32
C. 64
D. 96
E. 128


Solution:

Without any restrictions, the number of ways to choose 3 balls from 9 is 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84.

Let’s calculate the number of ways we don’t select a black ball. Since there are 6 non-black balls, we can choose any 3 of them in 6C3 = (6 x 5 x 4) / (3 x 2) = 120/6 = 20 ways.

In summary, there are 84 different selections of 3 balls (if there are no restrictions), and 20 of them consist of selecting no black balls. Therefore, there must be 84 - 20 = 64 different selection of 3 balls with at least one black ball.

Answer: C

What goes wrong when I say lets keep one black ball aside as we need "at least one black ball" and then select 2 balls out of 8. How am I counting incorrectly
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Bunuel
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

A. 16
B. 32
C. 64
D. 96
E. 128


Solution:

Without any restrictions, the number of ways to choose 3 balls from 9 is 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84.

Let’s calculate the number of ways we don’t select a black ball. Since there are 6 non-black balls, we can choose any 3 of them in 6C3 = (6 x 5 x 4) / (3 x 2) = 120/6 = 20 ways.

In summary, there are 84 different selections of 3 balls (if there are no restrictions), and 20 of them consist of selecting no black balls. Therefore, there must be 84 - 20 = 64 different selection of 3 balls with at least one black ball.

Answer: C

What goes wrong when I say lets keep one black ball aside as we need "at least one black ball" and then select 2 balls out of 8. How am I counting incorrectly

You under estimate the count of black balls that way, first you could've picked 1 or 2 black balls from 3 black balls, now since you have "fixed" one black ball, you only have 2 black balls to pick from.

CaseCorrect CountIncorrect Count (Underestimation)
1 black, 2 non-black3C1*6C2 = 456C2 = 15
2 black, 1 non-black3C2*6C1 = 182C1*6C1 = 12
3 black3C3 = 12C2 = 1
Total6428
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Total Balls = 9 (WWBBBRRRR)

Ways to choose ANY 3 balls from 9 (without defining the color) = 9c3 = 84

Ways to NOT Pick any black ball = 6c3 = 20

Ways to choose at least one black ball = 84 - 20 = 64.

Ans C
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Is the answer C (64) based on the assumption that the balls of the same color are distinct/differentiated?
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EasyGMATPrep
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

A. 16
B. 32
C. 64
D. 96
E. 128

Is the answer C (64) based on the assumption that the balls of the same color are distinct/differentiated?

We don’t care which black or white balls are in the selection as long as at least one black ball is included in the draw. However, the balls are still physically distinct objects, so when calculating the number of ways to select, say, 3 balls out of 9, we still do 9C3. This is because when reaching for the balls in the box, there are still 9 distinct objects, so this gives us the number of ways to pick 3 distinct objects from 9.
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