Let $\alpha=\frac{(4 !) !}{(4 !)^{3 !}}$ and $\beta=\frac{(5 !) !}{(5 !)^{4 !}}$. Then :
Solution
<p>$$\begin{aligned}
& \alpha=\frac{(4 !) !}{(4 !)^{3 !}}, \beta=\frac{(5 !) !}{(5 !)^{4 !}} \\
& \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}}
\end{aligned}$$</p>
<p>Let 24 distinct objects are divided into 6 groups of 4 objects in each group.</p>
<p>No. of ways of formation of group $=\frac{24 !}{(4 !)^6 .6 !} \in \mathrm{N}$<?p>
<p>Similarly,</p>
<p>Let 120 distinct objects are divided into 24 groups of 5 objects in each group.</p>
<p>No. of ways of formation of groups</p>
<p>$=\frac{(120) !}{(5 !)^{24} \cdot 24 !} \in \mathrm{N}$</p>
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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