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JEE Mains · Maths · STD 11 - 6. permutation and combination

Let \(\alpha=\frac{(4 !) !}{(4 !)^{3 !}}\) and \(\beta=\frac{(5 !) !}{(5 !)^{4 !}}\). Then :

  1. A \(\alpha \in \mathrm{N}\) and \(\beta \notin \mathrm{N}\)
  2. B \(\alpha \notin \mathrm{N}\) and \(\beta \in \mathrm{N}\)
  3. C \(\alpha \in \mathrm{N}\) and \(\beta \in \mathrm{N}\)
  4. D \(\alpha \notin \mathrm{N}\) and \(\beta \notin \mathrm{N}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\alpha \in \mathrm{N}\) and \(\beta \in \mathrm{N}\)

Step-by-step Solution

Detailed explanation

\(\alpha=\frac{(4 !) !}{(4 !)^{3 !}}, \beta=\frac{(5 !) !}{(5 !)^{4 !}} \) \( \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}}\) Let \(24\) distinct objects are divided into \(6\) groups of \(4\) objects in each group. No. of ways of formation of group…