Permutation Combination Question 3
Question 3 - 2024 (27 Jan Shift 2)
Let $\alpha=\frac{(4 !) !}{(4 !)^{3 !}}$ and $\beta=\frac{(5 !) !}{(5 !)^{4 !}}$. Then :
(1) $\alpha \in \mathrm{N}$ and $\beta \notin \mathrm{N}$
(2) $\alpha \notin \mathrm{N}$ and $\beta \in \mathrm{N}$
(3) $\alpha \in \mathrm{N}$ and $\beta \in \mathrm{N}$
(4) $\alpha \notin \mathrm{N}$ and $\beta \notin \mathrm{N}$
Show Answer
Answer (3)
Solution
$\alpha=\frac{(4 !) !}{(4 !)^{3 !}}, \beta=\frac{(5 !) !}{(5 !)^{4 !}}$
$\alpha=\frac{(24) !}{(4 !)^{6}}, \beta=\frac{(120) !}{(5 !)^{24}}$
Let 24 distinct objects are divided into 6 groups of 4 objects in each group.
No. of ways of formation of group $=\frac{24 !}{(4 !)^{6} .6 !} \in \mathrm{N}$
Similarly,
Let 120 distinct objects are divided into 24 groups of 5 objects in each group.
No. of ways of formation of groups
$=\frac{(120) !}{(5 !)^{24} .24 !} \in \mathrm{N}$
