Question

$\mathrm{S}=\{1,2,3\dots ,20\}$ is to be partitioned into four sets $\mathrm{A},\mathrm{B},\hspace{0.17em}\mathrm{C}$ and $\mathrm{D}$of equal size. The number of ways it can be done, is equal to $?$

• A. $\frac{20!}{(4!\mathrm{x}5!)}$
• B. $\frac{20!}{{\left(4\right)}^5}$
• C. $\frac{20!}{{(5!)}^4}$
• D. $\frac{20!}{{(4!)}^5}$

Answer 1

The correct option is C

$\frac{20!}{{(5!)}^4}$

Finding the number of ways:

Given, $\mathrm{S}=\{1,2,3\dots ,20\}\to 4$ sets
Let's consider $5$ elements per set. The number of ways of partitioning is as follows
To select $5$ elements for set $\mathrm{A}$from $20$ $\to {}^{20}\mathrm{C}_5.$
To select $5$ elements for set $\mathrm{B}$ from $15\to {}^{15}C_5$
To select $5$ elements for set $\mathrm{C}$ from $10\to {}^{10}c_5$
To select $5$ elements for the final set $5\to {}^{5}C_5$

So, The number of ways

$={}^{20}C_5\times {}^{15}C_5\times {}^{10}C_5{\times }^5{C}_5$

$$\begin{gathered} =\frac{\left(20\right)!}{\left(5\right)!\left(15\right)!}\times \frac{\left(15\right)!}{\left(10\right)!\left(5\right)!}\times \frac{\left(10\right)!}{\left(5\right)!\left(5\right)!}\times 1 \\ =\frac{\left(20\right)!}{{\left[\left(5\right)!\right]}^4} \end{gathered}$$

Therefore, the number of ways it can be done is $\frac{20!}{{(5!)}^4}$

Hence, option (C) is the correct answer.