Question

If $S = \left\{ {1,2,3,....,20} \right\}$ is to be partitioned into four sets A, B, C and D of equal size. The number of ways it can be done, is equal to?
$\eqalign{
  & 1)\dfrac{{20!}}{{\left( {4! \times 5!} \right)}} \cr
  & 2)\dfrac{{20!}}{{{4^5}}} \cr
  & 3)\dfrac{{20!}}{{{{\left( {5!} \right)}^4}}} \cr
  & 4)\dfrac{{20!}}{{{{\left( {4!} \right)}^5}}} \cr} $

Answer 1

Hint: In the given question, there is a set with numbers from $1$ to $20$. This given set has to be partitioned into $4$ other sets and they must have equal terms. Therefore, we understand that there are going to be $5$ numbers in each set. Now, to figure the number of ways to do it, we are going to use combinations since it is only to determine the number of ways and no particular order is given.
The formula for combination is as follows:
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where,
$n$is the total number of items in the set
$r$ is the number of the items that are selected from the set

Complete step-by-step answer:
The given set is, $S = \left\{ {1,2,3,....,20} \right\}$
The given set consists of $20$ items. So, in order to put them into $4$ different sets with equal size, we will place $5$ items in each set.
Using the above-mentioned formula, we can determine the number of ways in which we can partition it equally.
That is, ${ = ^{20}}{C_5}{ \times ^{15}}{C_5}{ \times ^{10}}{C_5}{ \times ^5}{C_5}$
Here, the total number of items for each set will always remain $5$
But the number of items we need to choose will keep decreasing by $5$ as we choose $5$ from each of the given sets.
Therefore,
$ = \dfrac{{20!}}{{\left( {20 - 5} \right)! \times 5!}} \times \dfrac{{15!}}{{\left( {15 - 5} \right)! \times 5!}} \times \dfrac{{10!}}{{\left( {10 - 5} \right)! \times 5!}} \times \dfrac{{5!}}{{5!}}$
Now, by further simplification,
$ = \dfrac{{20!}}{{15! \times 5!}} \times \dfrac{{15!}}{{10! \times 5!}} \times \dfrac{{10!}}{{5! \times 5!}} \times 1$
In the above expression, only $10!$and $15!$cancel out.
Therefore, we will be left with $\dfrac{{20!}}{{{{\left( {5!} \right)}^4}}}$which is the final answer.
Hence, option (3) is the correct answer.
So, the correct answer is “Option 3”.

Note: The given question asks for the number of ways so, choose Combination. The options are in terms of factorials, so we need not find out the values of them. Note that, we cannot cancel out the factorials unless they are equal, we cannot use normal division to simplify them.