JEE Mathematics 2024 Question with Solution

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

Find: Whether $\alpha$ and $\beta$ belong to $N$.

The solution provided is unrelated to this question and discusses vector collinearity with a final value of $35$, so it cannot be used to derive the answer for this factorial question.

Using the given expressions directly:

$$4! = 24,$$

so

$$\alpha = \frac{24!}{(24^3)!} = \frac{24!}{13824!}.$$

Since the numerator factorial is of a much smaller positive integer than the denominator factorial, $\alpha$ is a positive fraction and hence $\alpha \notin N$.

Also,

$$5! = 120,$$

so

$$\beta = \frac{120!}{(120^4)!} = \frac{120!}{207360000!}.$$

Again, this is a positive fraction less than $1$, hence $\beta \notin N$.

Therefore, both $\alpha$ and $\beta$ are not natural numbers. The correct option is D.

However, the answer key marks option C. This creates a source discrepancy. Based on the provided answer field, the recorded answer is C, but mathematically the expressions as written support D.