Question

Find the condition when ${}^{n}{{P}_{r}}$ and ${}^{n}{{C}_{r}}$ are equal:

& A.n=r \\\ & B.n=r+1 \\\ & C.r=1 \\\ & D.n=r-1 \\\ \end{aligned}$$

Answer 1

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

& \Rightarrow r!=1 \\\ & \Rightarrow r=1 \\\ \end{aligned}$$ $$\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.