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JEE Advanced · Mathematics · 4. P&C

A group of 9 students, \(\mathrm{s}_1, \mathrm{~s}_2, \ldots \ldots, \mathrm{s}_9\), is to be divided to from three teams \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\) of sizes 2,3 , and 4 , respectively. Suppose that \(s_1\) cannot be selected for the team \(\mathrm{X}\), and \(\mathrm{s}_2\) cannot be selected for the team Y. Then the number of ways to from such teams, is

  1. A 665
  2. B 401
  3. C 650
  4. D 140
Verified Solution

Answer & Solution

Correct Answer

(A) 665

Step-by-step Solution

Detailed explanation

\(\begin{array}{ccc}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 2 & 3 & 4 \end{array}\)
C-i) when \(x\) does not contain \(\mathrm{S}_1\), but contains \(\mathrm{S}_2\)
\(\underset{\text{(for }x)}{{ }^7 C_1} \times \underset{\text{(for }y,z)}{\frac{7!}{3!4!}}=245\)
C-ii) When \(\mathrm{x}\) does not contain \(\mathrm{S}_1, \mathrm{~S}_2\) and \(\mathrm{y}\) does not contain \(\mathrm{S}_2\) i.e. \(\underset{\text{(for }x)}{{ }^{7} \mathrm{C}_2} \times \underset{\text{(for }y,z)}{\frac{6!}{3!3!}}=420\)
so total No. of ways \(665\)
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