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 8 of 9.
Gauss said:
10 years ago
Can anybody tell me why the answer is not 3C1 x 8C2 = 84?
Apluv said:
10 years ago
Answer is 64. If the colors are different then balls are not identical.
Naveen said:
10 years ago
@Pavel Sain: Since the ball of the same colour are identical, you should do ((3C1/3!) * (2C1/2!) * (4C1/4!)) etc.
So either you use the formula or make the individual combinations, the answer will be 6.
So either you use the formula or make the individual combinations, the answer will be 6.
Bhuvi said:
9 years ago
It can be the balls are of same colours but different shades.
In my book, RD Sharma's objective mathematics it is done like 9C3 - 6c 3 which created doubt in my mind he took all the balls together which is not possible in case of identical balls. As there is only 1 way to select 2 balls from 20 identical balls.
In my book, RD Sharma's objective mathematics it is done like 9C3 - 6c 3 which created doubt in my mind he took all the balls together which is not possible in case of identical balls. As there is only 1 way to select 2 balls from 20 identical balls.
Himanshu Kulkarni said:
10 years ago
One black ball from 3 is 3C1 and any 2 balls from remaining 8 balls
i.e. 8C2.
So, why 3C1 x 8C2 does not match the solution?
i.e. 8C2.
So, why 3C1 x 8C2 does not match the solution?
Miles said:
1 decade ago
Consider the case where only one black ball is chosen. It could be chosen first second or third so that's 3 possibilities. As for the other two balls, they could be red or white. It could be red then white, white then red, white then white, or red then red, regardless of when the black ball is chosen.
So that's 4 possibilities. So, with only one black ball, that would be 12 possibilities (brw, bwr, brr, bww, rbw, wbr, rbr, wbw, rwb, wrb, rrb, wwb) but the solutions says there are 45 possibilities for this condition. Am I missing something or what?
So that's 4 possibilities. So, with only one black ball, that would be 12 possibilities (brw, bwr, brr, bww, rbw, wbr, rbr, wbw, rwb, wrb, rrb, wwb) but the solutions says there are 45 possibilities for this condition. Am I missing something or what?
Sanjay said:
1 decade ago
Why not 3C1*8c2?
Prasoon garg said:
1 decade ago
I agree with @Mani as it is not mentioned in the question that all the balls are non-identical.
Sundar said:
1 decade ago
We should not take 3 black balls. Then why final declaration 3 ball added?
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.
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