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 ___________.

Solution

Maths sir · 4 steps

At most three red is everything except all four red, so count the total and remove that one case.

1. Total draws. There are 5 + 4 + 3 = 12 marbles. Each marble counts as different, so any 4 of them can be chosen in C(12, 4) = 495 ways.
2. The bad case. At most three red fails only when all 4 marbles are red. Choosing 4 of the 5 red marbles gives C(5, 4) = 5 ways.

   C(12, 4) - C(5, 4) = 495 - 5 = 490

3. Direct check. Counting by the number of red marbles: 0 red gives C(7, 4) = 35, 1 red gives 5 × 35 = 175, 2 red gives 10 × 21 = 210, 3 red gives 10 × 7 = 70. These add to 490.
4. So. The number of ways is 490.

The common slip: 455. Leaving out the case with no red marble (C(7, 4) = 35 ways) gives 455, not 490. At most three red includes zero red.

The idea to keep. When a condition says at most or at least, first ask what it leaves out. Often the total minus one or two bad cases is far quicker than adding every allowed case.

Answer from the NTA final answer key. Solution by PrepWiser.