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:
82 comments Page 9 of 9.
Mukesh said:
1 decade ago
Here you can draw one black ball from x number of black balls in only one way, similarly 2 red balls from x numbers of red balls also in one way for xC2.
So, (3 Black) + (2 Black AND (1 Red OR 1 White)) + (1 Black AND ((2 white OR 2 Red OR (1 White AND 1 Red))).
-> 1 + 1 * 2 + 1 * (1+1+1).
-> 1 + 2 + 3.
-> 6.
Which I believe should be answer, any corrections or clarification most appreciated.
So, (3 Black) + (2 Black AND (1 Red OR 1 White)) + (1 Black AND ((2 white OR 2 Red OR (1 White AND 1 Red))).
-> 1 + 1 * 2 + 1 * (1+1+1).
-> 1 + 2 + 3.
-> 6.
Which I believe should be answer, any corrections or clarification most appreciated.
Sasi said:
1 decade ago
I have a doubt. In last question you have given nC(n-r). Why This conditions is not suitable for this?
In this question also why can't we do this as:
= (3C1 x 6C(6-2)) + (3C(3-2) x 6C1) + (3C3).
In this question also why can't we do this as:
= (3C1 x 6C(6-2)) + (3C(3-2) x 6C1) + (3C3).
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