Question

Find the condition when \[{}^{n}{{P}_{r}}\] and \[{}^{n}{{C}_{r}}\] are equal:
\[\begin{align}
  & A.n=r \\
 & B.n=r+1 \\
 & C.r=1 \\
 & D.n=r-1 \\
\end{align}\]

Answer 1

Hint: In order to calculate when \[{}^{n}{{P}_{r}}\] and \[{}^{n}{{C}_{r}}\]would be equal, firstly we have to evaluate or expand the formula of both \[{}^{n}{{P}_{r}}\] and \[{}^{n}{{C}_{r}}\]. Then we must equate both of the expansions and cancel out the common terms and obtain the final result. This would be our required result.

Complete step by step answer:
Now let us learn about the permutation and combinations. Permutation is nothing but arranging the members in some sequence or following a particular order whereas combination is a collection where it does not follow any sought of order or sequence. There are two types of permutations and combinations as well. The two types of permutations are: permutation with repetition and permutation without repetition. There are two types of combinations like permutations. They are: combination with repetition and combination without repetition.
Now let us find when\[{}^{n}{{P}_{r}}\] and \[{}^{n}{{C}_{r}}\] would be equal.
In order to find, firstly let us expand \[{}^{n}{{P}_{r}}\] and \[{}^{n}{{C}_{r}}\]
\[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\] and
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Now let us equate both of the expansions.
\[\Rightarrow \dfrac{n!}{\left( n-r \right)!}=\dfrac{n!}{r!\left( n-r \right)!}\]
Upon cancelling out the common terms, we will be left with
\[\Rightarrow 1=\dfrac{1}{r!}\]
Upon cross multiplying, we get
\[\begin{align}
  & \Rightarrow r!=1 \\
 & \Rightarrow r=1 \\
\end{align}\]
\[\therefore \] \[{}^{n}{{P}_{r}}\] and \[{}^{n}{{C}_{r}}\] are equal when \[r=1\].

So, the correct answer is “Option C”.

Note: We can apply the permutation and combinations in our everyday life. We can apply permutation in finding out the captain of a team, or picking two favourite colours. We can use combinations in picking up three members for a team from a group or picking three winners and many more.