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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

There are $n$ letters and $n$ addressed envelopes. The probability that all the letters are not kept in the right envelope is:

In how many ways can a pack of $52$ cards be divided into four equal groups?

Let $A = \{x_1, x_2, x_3, x_4\}$ and $B = \{y_1, y_2, y_3, y_4\}$. $A$ function $f: A \to B$ is defined. The number of one-one functions such that $f(x_i) \neq y_i$ for $i = 1, 2, 3, 4$ is equal to:

Choose the correct number of ways in which $15$ different books can be divided into $5$ heaps of equal number of books.

If $5$ letters are to be placed in $5$ addressed envelopes,then the probability that at least one letter is placed in the wrongly addressed envelope is

Vedclass Test Series

Exam Paper Generator

Online Exam Module