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 3 of 9.
Krish said:
1 decade ago
3c1*8c2 = 84?
Why not? Atleast one black ball is there right? And from remainder we take 2 balls. What is wrong in this?
Why not? Atleast one black ball is there right? And from remainder we take 2 balls. What is wrong in this?
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).
Bbbb said:
1 decade ago
@ Saireshwanth and Saurabh.
Could you explain how you get 64? Because I got the answer as 73.
Could you explain how you get 64? Because I got the answer as 73.
(1)
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.
Jumenal said:
1 decade ago
How you taken 6 in combination?
(1)
Tanya said:
1 decade ago
Can't it be like 1 black ball is chosen from 3 black balls and then remaining 2 are chosen from 2 white, 4 red and remaining 2 black balls i.e 3.
= 3C1*8C2.
= 3C1*8C2.
(1)
Pavel Sain said:
1 decade ago
May be we can take (1B+1W+1R) OR (1B+2W) OR (1B+2R) OR (2B+1W) OR (2B+1R) OR all (3B).
So required no = (3C1*2C1*4C1)+(3C1*2C2)+(3C1*4C2)+(3C2*2C1)+(3C2*4C1)+3C3.
= (3*2*4)+(3*1)+(3*6)+(3*2)+(3*4)+1.
= 24+3+18+6+12+1 = 64.
So required no = (3C1*2C1*4C1)+(3C1*2C2)+(3C1*4C2)+(3C2*2C1)+(3C2*4C1)+3C3.
= (3*2*4)+(3*1)+(3*6)+(3*2)+(3*4)+1.
= 24+3+18+6+12+1 = 64.
Sundar said:
1 decade ago
We should not take 3 black balls. Then why final declaration 3 ball added?
Prasoon garg said:
10 years ago
I agree with @Mani as it is not mentioned in the question that all the balls are non-identical.
Sanjay said:
10 years ago
Why not 3C1*8c2?
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