Let alpha = frac{(4!)!}{(4!)^{3!}} and beta = | vedclass

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(C) Given $\alpha = \frac{(4!)!}{(4!)^{3!}} = \frac{24!}{(24)^6}$ and $\beta = \frac{(5!)!}{(5!)^{4!}} = \frac{120!}{(120)^{24}}$.The number of ways t

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$\alpha \in N$ and $\beta \in N$

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(C) Given $\alpha = \frac{(4!)!}{(4!)^{3!}} = \frac{24!}{(24)^6}$ and $\beta = \frac{(5!)!}{(5!)^{4!}} = \frac{120!}{(120)^{24}}$.The number of ways to divide $n$ distinct objects into $k$ groups of size $m$ each (where $n = km$) is given by $\frac{n!}{(m!)^k \cdot k!}$.For $\alpha$,we have $n=24, m=4, k=6$. The number of ways is $\frac{24!}{(4!)^6 \cdot 6!} = K_1$,where $K_1 \in N$.Thus,$\alpha = K_1 \cdot 6!$. Since $K_1$ and $6!$ are integers,$\alpha \in N$.For $\beta$,we have $n=120, m=5, k=24$. The number of ways is $\frac{120!}{(5!)^{24} \cdot 24!} = K_2$,where $K_2 \in N$.Thus,$\beta = K_2 \cdot 24!$. Since $K_2$ and $24!$ are integers,$\beta \in N$.Therefore,both $\alpha$ and $\beta$ are natural numbers.

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

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