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?
Answer: Option
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) | |||||||||||||
|
||||||||||||||
| = (45 + 18 + 1) | ||||||||||||||
| = 64. |
Discussion:
81 comments Page 9 of 9.
Mani said:
1 decade ago
I have a doubt in this.
Its not mentioned in the question that all the balls are non-identical. It is said '3 black balls', so all the 3 black balls are identical.
The required no. of ways = (1 black and 2 white) or (1 black and 2 red) or (1 black and 1 white and 1 red) or (2 black and 1 white) or (2 black and 1 red) or (3 black)
= 1+1+1+1+1+1
= 6ways.
Can anyone please confirm my answer?
Thank u :)
Its not mentioned in the question that all the balls are non-identical. It is said '3 black balls', so all the 3 black balls are identical.
The required no. of ways = (1 black and 2 white) or (1 black and 2 red) or (1 black and 1 white and 1 red) or (2 black and 1 white) or (2 black and 1 red) or (3 black)
= 1+1+1+1+1+1
= 6ways.
Can anyone please confirm my answer?
Thank u :)
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