Solveeit Logo
Login

Question

Question: The value of \(nP_{r}\) is equal to....

The value of nPrnP_{r} is equal to.

A

n−1Pr+rn−1Pr−1n - 1P_{r} + r^{n - 1}P_{r - 1}

B

n. 6mun−1Pr+n−1⥂Pr−1n.\mspace{6mu}^{n - 1}P_{r} +^{n - 1} ⥂ P_{r - 1}

C

n(n−1Pr+n−1⥂Pr−1)n(^{n - 1}P_{r} +^{n - 1} ⥂ P_{r - 1})

D

n−1Pr−1+n−1Prn - 1P_{r - 1} +^{n - 1}P_{r}

Answer

n−1Pr+rn−1Pr−1n - 1P_{r} + r^{n - 1}P_{r - 1}

Explanation

Solution

n−1Pr+r.n−1⥂Pr−1n - 1P_{r} + r.^{n - 1} ⥂ P_{r - 1}

=(n−1)!(n−1−r)!+r(n−1)!(n−r)!= \frac{(n - 1)!}{(n - 1 - r)!} + r\frac{(n - 1)!}{(n - r)!} (∵nPr=n!(n−r)!)\left( \because^{n}P_{r} = \frac{n!}{(n - r)!} \right)

= (n−1)!(n−1−r)!{1+r.1n−r}\frac{(n - 1)!}{(n - 1 - r)!}\left\{ 1 + r.\frac{1}{n - r} \right\}

= (n−1)!(n−1−r)!(n−r)!(nn−r)=n!(n−r)!=nPr\frac{(n - 1)!}{(n - 1 - r)!(n - r)!}\left( \frac{n}{n - r} \right) = \frac{n!}{(n - r)!} =^{n}P_{r}.

Aliter : We know that

n−1Cr+n−1Cr−1=nCrn - 1C_{r} +^{n - 1}C_{r - 1} =^{n}C_{r}

⇒ n−1Prr!+n−1Pr−1(r−1)!=n⥂Prr!\frac{n - 1P_{r}}{r!} + \frac{n - 1P_{r - 1}}{(r - 1)!} = \frac{n ⥂ P_{r}}{r!} ⇒ n−1Pr+r.n−1Pr−1=nPrn - 1P_{r} + r.^{n - 1}P_{r - 1} =^{n}P_{r}.