Question

If two sets \[A\] and \[B\] are having 99 elements in common, then what is the number of elements common to each of the sets \[A \times B\] and \[B \times A\]?
A. \[{2^{99}}\]
B. \[{99^2}\]
C. 100
D. 18
E. 9

Answer 1

Hint: In the given question the common elements is 99. So, we will find the number of ways to select the first element of the ordered pairs and the number of ways to select the second element of the ordered pairs. Then multiply the ways.

Formula used:
\[A \times B\] is an ordered pair such that \[\left\{ {\left( {a,b} \right):a \in A\,{\rm{and}}\,b \in B} \right\}\].

Complete step by step solution:
Given that, two sets \[A\] and \[B\] are having 99 elements in common.
Suppose the common elements of set \[A\] and set \[B\] are \[{n_1},{n_2},{n_3}, \cdots ,{n_{99}}\].
The number of ways to select the first element of the ordered pairs of \[A \times B\] from the 99 common elements is \[{}^{99}{C_1}\].
Now apply the combination formula.
\[{}^{99}{C_1} = \dfrac{{99!}}{{\left( {99 - 1} \right)!1!}}\]
\[ \Rightarrow {}^{99}{C_1} = \dfrac{{99 \times 98!}}{{98!1!}}\]
Cancel out \[98!\] from the denominator and numerator.
\[ \Rightarrow {}^{99}{C_1} = 99\]
Now calculate the number of ways to select the second element of the ordered pairs of \[A \times B\].
The number of ways to select the second element of the ordered pairs of \[A \times B\] from the 99 common elements is \[{}^{99}{C_1}\].
Now apply the combination formula.
\[{}^{99}{C_1} = \dfrac{{99!}}{{\left( {99 - 1} \right)!1!}}\]
\[ \Rightarrow {}^{99}{C_1} = \dfrac{{99 \times 98!}}{{98!1!}}\]
Cancel out \[98!\] from the denominator and numerator.
\[ \Rightarrow {}^{99}{C_1} = 99\]
So, the total number of ways to make ordered pairs of \[A \times B\] from the common elements is \[99 \times 99 = {99^2}\].
Similar way we can get the number of ways to make ordered pairs of \[B \times A\] from the common elements is \[99 \times 99 = {99^2}\].
So, the common elements of the sets \[A \times B\] and \[B \times A\] is \[{99^2}\].
Hence option B is the correct option.

Note: We calculate the common elements of the sets \[A \times B\] and \[B \times A\] by using the formula \[n\left( {\left( {A \times B} \right) \cap \left( {B \times A} \right)} \right) = n {\left( {A \cap B} \right) \times n\left( {B \cap A} \right)} \]. By using the formula, we will \[n\left( {\left( {A \times B} \right) \cap \left( {B \times A} \right)} \right) = n {\left( {A \cap B} \right) \times n\left( {B \cap A} \right)} = 99 \times 99 = {99^2}\]. Thus, the common elements of the sets \[A \times B\] and \[B \times A\] is \[{99^2}\].