Permutations & Combinations
Selection with Restrictions
Grade 11
Question:
<p>An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Find the number of ways in which 4 marbles can be drawn so that at most three of them are red.</p>
Step-by-Step Solution
Key Concept: Use complementary counting: subtract the unwanted case (all 4 red) from total possible ways to draw 4 marbles.
<p><strong>Step 1:</strong> Total number of ways to draw 4 marbles from 12 marbles = \(\binom{12}{4}\)</p><p><strong>Step 2:</strong> Number of ways to draw 4 marbles such that all are red = \(\binom{5}{4}\)</p><p><strong>Step 3:</strong> Number of ways such that at most three are red = Total ways − Ways with all red marbles</p><p>\[\binom{12}{4} - \binom{5}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2} - 5\]</p><p>\[= 11 \times 5 \times 9 - 5 = 495 - 5 = 490\]</p><p>∴ Answer is 490.</p>
Correct Answer: 490