JEE Mains · Maths · STD 11 - 6. permutation and combination
Let \(\alpha=\frac{(4 !) !}{(4 !)^{3 !}}\) and \(\beta=\frac{(5 !) !}{(5 !)^{4 !}}\). Then :
- A \(\alpha \in \mathrm{N}\) and \(\beta \notin \mathrm{N}\)
- B \(\alpha \notin \mathrm{N}\) and \(\beta \in \mathrm{N}\)
- C \(\alpha \in \mathrm{N}\) and \(\beta \in \mathrm{N}\)
- D \(\alpha \notin \mathrm{N}\) and \(\beta \notin \mathrm{N}\)
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…
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