<p>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 ………. (JEE Main 2020)</p>
Step-by-Step Solution
Key Concept: Break down 'at most 3 red' into cases (0, 1, 2, 3 red marbles) and use combinations to count each case.
<p><strong>Solution:</strong></p><p>Total marbles: 5 red + 4 black + 3 white = 12 marbles</p><p>Drawing 4 marbles such that at most 3 are red means: 0, 1, 2, or 3 red marbles.</p><p>Case 1: 0 red, 4 non-red: \(\binom{5}{0}\binom{7}{4} = 1 \times 35 = 35\)</p><p>Case 2: 1 red, 3 non-red: \(\binom{5}{1}\binom{7}{3} = 5 \times 35 = 175\)</p><p>Case 3: 2 red, 2 non-red: \(\binom{5}{2}\binom{7}{2} = 10 \times 21 = 210\)</p><p>Case 4: 3 red, 1 non-red: \(\binom{5}{3}\binom{7}{1} = 10 \times 7 = 70\)</p><p>Total = 35 + 175 + 210 + 70 = 490</p>
Correct Answer: 490