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

(C) $\alpha \in N$ and $\beta \in N$

(A) $\alpha \in N$ and $\beta \notin N$

(B) $\alpha \notin \mathrm{N}$ and $\beta \in \mathrm{N}$

(D) $\alpha \notin \mathrm{N}$ and $\beta \notin \mathrm{N}$

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Permutations and Combinations

Let

α = \frac{( 4 !)!}{( 4 ! ) ^{3!}}

and

β = \frac{( 5 !)!}{( 5 ! ) ^{4!}}

. Then :

(A) $α \in N$ and $β \in / N$
(B) $α \in / N$ and $β \in N$
(C) $α \in N$ and $β \in N$ ✓ Correct
(D) $α \in / N$ and $β \in / N$

Answer: (C)

Solution

α = \frac{( 4 !)!}{( 4 ! ) ^{3!}} = \frac{24 !}{2 4 ^{6}} = \frac{24 !}{( 4 ! ) ^{6}}

.

This counts the number of ways to split 24 distinct objects into 6 labelled groups of 4 objects each — a count of arrangements, hence a positive integer. So

α \in N

β = \frac{( 5 !)!}{( 5 ! ) ^{4!}} = \frac{120 !}{12 0 ^{24}} = \frac{120 !}{( 5 ! ) ^{24}}

.

Similarly, this counts the ways to split 120 distinct objects into 24 labelled groups of 5 objects each, also a positive integer:

β \in N

.

Both

α \in N

and

β \in N

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Permutations and Combinations

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