Let α = \; {(4!)!}{(4!)³!}} and β = \; {(5!)!}{(5!)^{4!}}. Then : [27-Jan-2024 Shift 2]

Correct Answer is : α∈Nα {N}α∈N and β∈Nβ {N}β∈N

Correct Answer is : α∈Nα {N}α∈N and β∈Nβ {N}β∈N

Test Index

Permutations and Combinations Part 2

Section: Mathematics

Question : 64 of 120

Marks: +1, -0

Let

\alpha = \; \frac{(4!)!}{(4!)^{3!}}

\beta = \; \frac{(5!)!}{(5!)^{4!}}

Solution: 👈: Video Solution

\; \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

= \; \frac{24!}{(4!)^6 \cdot 6!} \in \mathbb{N}

Similarly,

Let 120 distinct objects are divided into 24 groups of 5 objects in each group.

No. of ways of formation of groups

= \; \frac{(120)!}{(5!)^{24} \cdot 24!} \in \mathbb{N}

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