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Question: Prove that \({}^{\text{n}}{{\text{C}}_{\text{r}}}\) \( \times \) r!= \({}^{\text{n}}{{\text{P}}_{\te...

Prove that nCr{}^{\text{n}}{{\text{C}}_{\text{r}}} ×\times r!= nPr{}^{\text{n}}{{\text{P}}_{\text{r}}}

Explanation

Solution

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 4C2{}^{\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 4P2{}^{\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. nPr{}^{\text{n}}{{\text{P}}_{\text{r}}} in this case as discuss below.

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