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:
79 comments Page 1 of 8.
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.
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.
(13)
Tammunur Joshna said:
2 years 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.
= 3*2*4 + 3*4 + 3*2 + 6*3 + 3 +1.
= 64.
(8)
Aditya Paras said:
5 months ago
@All.
Please Mention distinct balls, usually the same colour is used for similar objects
Please Mention distinct balls, usually the same colour is used for similar objects
(3)
AkashSingh said:
6 years ago
1 Black + 1 White + 1 Red.
2 Black + 1 White.
1 Black + 2 White.
2 Black + 1 Red.
2 Red + 1 Black.
3 Black.
2 Black + 1 White.
1 Black + 2 White.
2 Black + 1 Red.
2 Red + 1 Black.
3 Black.
(1)
Anil said:
8 years ago
Can anybody tell me why the answer is not 3C1 x 8C2 = 84?
(1)
Supratik said:
7 years ago
Why not 3C1 x 8C2=84 ?
(1)
Ashu Pathak said:
7 years ago
Why not 3c1 * 8c2=84?
(1)
Tanya said:
10 years 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)
Jumenal said:
1 decade ago
How you taken 6 in combination?
(1)
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)
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