JEE Mains · Maths · STD 11 - 4.1 complex nubers
A box contains \(5\) blue, \(6\) yellow and \(4\) red balls. The number of ways, of drawing \(8\) balls containing at least two balls of each colour, is :
- A \(4100\)
- B \(4140\)
- C \(4230\)
- D \(4290\)
Answer & Solution
Correct Answer
(A) \(4100\)
Step-by-step Solution
Detailed explanation
Let \(B\), \(Y\), and \(R\) be the number of blue, yellow, and red balls drawn respectively. We are given \(B + Y + R = 8\) with \(B \ge 2\), \(Y \ge 2\), and \(R \ge 2\). The possible combinations for \((B, Y, R)\) are: Case 1: \((2, 2, 4)\) Case 2: \((2, 3, 3)\) Case 3:…
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