Question

Prove that ${}^{\text{n}}{{\text{C}}_{\text{r}}}$$ \times $ r!= ${}^{\text{n}}{{\text{P}}_{\text{r}}}$

Answer 1

Hint:In this question we use the theory of permutation and combination. So, before solving this question you need to first recall the basics formula of this chapter. For example, if we need to select two things out of four things. In this case, this can be done in ${}^{\text{4}}{{\text{C}}_{\text{2}}}$ =6 ways. And similarly, if we need to arrange two things out of four things then it will be done by ${}^{\text{4}}{{\text{P}}_{\text{2}}}$ ways. Here we first take LHS of the given expression and solve it until we will get the RHS i.e. ${}^{\text{n}}{{\text{P}}_{\text{r}}}$ in this case as discuss below.

Complete step-by-step solution:
We need to prove-
${}^{\text{n}}{{\text{C}}_{\text{r}}}$$ \times $ r! = ${}^{\text{n}}{{\text{P}}_{\text{r}}}$
Now, we take LHS of the given expressions-
LHS =${}^{\text{n}}{{\text{C}}_{\text{r}}}{ \times r! }$
As we know,
${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
And also,
${}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[(n - r)!]}}}}$
Now, put the value of ${}^{\text{n}}{{\text{C}}_{\text{r}}}$ in LHS of the given expression and we get
LHS =${}^{\text{n}}{{\text{C}}_{\text{r}}}$$ \times $ r!
= $\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$$ \times r!$
= $\dfrac{{{\text{n}}!}}{{{\text{[(n - r)}}!{\text{]}}}}$
= ${}^{\text{n}}{{\text{P}}_{\text{r}}}$
As we know, \[\left[ {{}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[(n - r)!]}}}}} \right]\]
Therefore, we verify that ${}^{\text{n}}{{\text{C}}_{\text{r}}}$$ \times $ r!=${}^{\text{n}}{{\text{P}}_{\text{r}}}$

Note: A permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collections, in such a way that the order of the objects does not matter. For example, suppose we have a set of three letters: A, B, and C. Each possible selection would be an example of a combination.