# Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 9)
9.
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?
32
48
64
96
None of these
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 + 3 x 2 x 6 + 1 2 x 1 2 x 1
= (45 + 18 + 1)
= 64.

Discussion:
78 comments Page 1 of 8.

Kabilan said:   2 months ago
Why all three drawn can not be black balls in these cases why is it not included?

Very simple, here we have to first find the 9c3 - 6c3.

Tammunur Joshna said:   1 year ago
3C1 * 2C1 * 4C1 + 3C2 * 4C1 + 3C2*2C1 + 4C2*3C1 + 2C2*3C1 + 3C3.
= 3*2*4 + 3*4 + 3*2 + 6*3 + 3 +1.
= 64.
(6)

Vish said:   2 years ago
We can also do it this way,

Atleast one black ball = total balls - No black ball
= 9c3 - 6c3.
9c3= 3 balls selected from the total no of balls,
6c3 - 3 balls taken from the total no of balls without any black which is 6(2 white + 3 red).

By calculating this, we will get the answer.
(12)

Emmy said:   2 years ago
I don't understand this, please. Can anyone help me out?

Ajle said:   2 years ago

Voldy said:   2 years ago
Why is the following method giving wrong answers:

3C1*9C1*9C1 = 243.
But 3C1 for 1 black ball and other 2 9C1s for 2 balls of any colour.
Can someone say, why this logic is wrong?

@All.

The Alternate way for the solution is:

All ways of picking 3 balls - All ways in which no black ball is picked = All ways in atleast 1 black ball is pick;

9c3 - 6c3 = 84 - 20 = 64.
(1)

Shubham said:   4 years ago
Total ways - no blackball = Atleast one black ball.
i.e,, 9C3-6C3.
(1)

Stige said:   4 years ago
(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

How to write this and (3C1 x 6C2) + (3C2 x 6C1)?