Question

The set S:\left\\{ {1,2,3,....,12} \right\\} is to be partitioned into three sets $A,B,C$ of equal size. Thus, $A \cup B \cup C = S,A \cap B = B \cap C = A \cap C = \phi$. The number of ways to partition$S$is
A. $\dfrac{{12!}}{{3!{{\left( {4!} \right)}^3}}}$
B. $\dfrac{{12!}}{{3!{{\left( {3!} \right)}^4}}}$
C. $\dfrac{{12!}}{{{{\left( {4!} \right)}^3}}}$
D. $\dfrac{{12!}}{{{{\left( {3!} \right)}^4}}}$

Answer 1

This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We need to know how to find a factorial value for a particular number. Also, we need to know how to expand sequence and cubic values into simple numbers. Also, we need to know the finding process of the required number of ways.

Complete step-by-step answer:
In this question, they are given,
S:\left\\{ {1,2,3,....,12} \right\\}
$A \cup B \cup C = S,A \cap B = B \cap C = A \cap C = \phi$
From the given question we get,
S:\left\\{ {1,2,3,....,12} \right\\} is to be partitioned into sets $A,B,C$ of equal size.
That is each set of four elements.
It is given in the question
$A \cap B = B \cap C = A \cap C = \phi$
So, in each set all the elements are different. So, the number of ways to partition is given below,
${}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4}$
Here, ${}^{12}{C_4} = \dfrac{{12!}}{{8! \times 4!}}$
${}^8{C_4} = \dfrac{{8!}}{{4! \times 4!}}$
${}^4{C_4} = 1$
By using these three values, we get
${}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = \dfrac{{12!}}{{8! \times 4!}} \times \dfrac{{8!}}{{4! \times 4!}} \times 1$
We can cancel the term$8!$in the numerator and the term$8!$ in the denominator. So, we get
{}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = \dfrac{{12!}}{{4!}} \times \dfrac{1}{{4! \times 4!}} \times 1$$$$ \to \left( 1 \right)
We know that,
$a \times a \times a$can also be written as ${a^3}$.
So, we get$4! \times 4! \times 4! = {\left( {4!} \right)^3}$
So, the equation$\left( 1 \right)$becomes,

$$\left( 1 \right) \to {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = \dfrac{{12!}}{{4!}} \times \dfrac{1}{{4! \times 4!}} \times 1 \\\ {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = \dfrac{{12!}}{{{{\left( {4!} \right)}^3}}} \\\$$

So, the final answer is,
The required number of ways$= \dfrac{{12!}}{{{{\left( {4!} \right)}^3}}}$
So, the correct answer is “Option C”.

Note: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. Note that if the same term is present in the numerator and the denominator we can cancel the term each other. Also, note that if the same term is multiplied thrice, we can write the expression in cubic form $n \times n \times n = {n^3}$.