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An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is _________.
Correct answer is '490'. Can you explain this answer?
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An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Th...
Required number of ways = Total - (All selected balls are red)
= 12C4 - 5C4
= 495 - 5 = 490
= 12C4 - 5C4
= 495 - 5 = 490
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An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Th...
To find the number of ways in which 4 marbles can be drawn from the given urn, we need to consider the different cases based on the number of red marbles drawn.
Case 1: Drawing 0 red marbles
In this case, we need to choose 4 marbles from the remaining 9 marbles (4 black + 3 white + 2 red). The number of ways to do this is given by the combination formula:
C(9, 4) = 9! / (4! * (9-4)!) = 126
Case 2: Drawing 1 red marble
In this case, we need to choose 1 red marble and 3 marbles from the remaining 9 marbles. The number of ways to do this is:
C(5, 1) * C(9, 3) = (5! / (1! * (5-1)!)) * (9! / (3! * (9-3)!)) = 5 * 84 = 420
Case 3: Drawing 2 red marbles
In this case, we need to choose 2 red marbles and 2 marbles from the remaining 9 marbles. The number of ways to do this is:
C(5, 2) * C(9, 2) = (5! / (2! * (5-2)!)) * (9! / (2! * (9-2)!)) = 10 * 36 = 360
Case 4: Drawing 3 red marbles
In this case, we need to choose 3 red marbles and 1 marble from the remaining 9 marbles. The number of ways to do this is:
C(5, 3) * C(9, 1) = (5! / (3! * (5-3)!)) * (9! / (1! * (9-1)!)) = 10 * 9 = 90
Total number of ways:
To find the total number of ways, we need to sum up the number of ways from all the cases:
126 + 420 + 360 + 90 = 996
But the question asks for the number of ways in which at most three of the marbles are red, so we need to subtract the case where all four marbles are red:
996 - 1 = 995
Therefore, the correct answer is 995, not 490 as stated in the question.
Case 1: Drawing 0 red marbles
In this case, we need to choose 4 marbles from the remaining 9 marbles (4 black + 3 white + 2 red). The number of ways to do this is given by the combination formula:
C(9, 4) = 9! / (4! * (9-4)!) = 126
Case 2: Drawing 1 red marble
In this case, we need to choose 1 red marble and 3 marbles from the remaining 9 marbles. The number of ways to do this is:
C(5, 1) * C(9, 3) = (5! / (1! * (5-1)!)) * (9! / (3! * (9-3)!)) = 5 * 84 = 420
Case 3: Drawing 2 red marbles
In this case, we need to choose 2 red marbles and 2 marbles from the remaining 9 marbles. The number of ways to do this is:
C(5, 2) * C(9, 2) = (5! / (2! * (5-2)!)) * (9! / (2! * (9-2)!)) = 10 * 36 = 360
Case 4: Drawing 3 red marbles
In this case, we need to choose 3 red marbles and 1 marble from the remaining 9 marbles. The number of ways to do this is:
C(5, 3) * C(9, 1) = (5! / (3! * (5-3)!)) * (9! / (1! * (9-1)!)) = 10 * 9 = 90
Total number of ways:
To find the total number of ways, we need to sum up the number of ways from all the cases:
126 + 420 + 360 + 90 = 996
But the question asks for the number of ways in which at most three of the marbles are red, so we need to subtract the case where all four marbles are red:
996 - 1 = 995
Therefore, the correct answer is 995, not 490 as stated in the question.
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An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is _________.Correct answer is '490'. Can you explain this answer?
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