Question

(a) Prove if P(n,r) = P(n-1,r) + r.P(n-1, r-1)​

Answer 1

We have to prove that,

$^n\!P_r\ =\ ^{n-1}\!P_r\ +\ r\cdot\ ^{n-1}\!P_{r-1}$

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First we write the three permutations in fractional form.

$^n\!P_r=\dfrac{n!}{(n-r)!}$

$^{n-1}\!P_r=\dfrac{(n-1)!}{(n-1-r)!}=\dfrac{(n-1)!}{(n-r-1)!}$

$^{n-1}\!P_{r-1}=\dfrac{(n-1)!}{(n-1-r+1)!}=\dfrac{(n-1)!}{(n-r)!}$

Since RHS has more terms, let me take the RHS first. So,

$\begin{aligned}&\textsf{RHS}\\ \\ \Longrightarrow\ \ &^{n-1}\!P_r+r\cdot\ ^{n-1}\!P_{r-1}\\ \\ \Longrightarrow\ \ &\frac{(n-1)!}{(n-r-1)!}+r\cdot \frac{(n-1)!}{(n-r)!}\\ \\ \Longrightarrow\ \ &\frac{(n-1)!}{(n-r-1)!}+\frac{r(n-1)!}{(n-r-1)!(n-r)}\\ \\ \Longrightarrow\ \ &\frac{(n-1)!}{(n-r-1)!}\left(1+\frac{r}{n-r}\right)\\ \\ \Longrightarrow\ \ &\frac{(n-1)!}{(n-r-1)!}\cdot \frac{n-r+r}{n-r}\\ \\ \Longrightarrow\ \ &\frac{(n-1)!}{(n-r-1)!}\cdot\frac{n}{n-r}\end{aligned}$

$\Longrightarrow\ \ \dfrac{(n-1)!\ n}{(n-r-1)!\ (n-r)}\\ \\ \\ \Longrightarrow\ \ \dfrac{n!}{(n-r)!}\\ \\ \\ \Longrightarrow\ \ ^n\!P_r\\ \\ \\ \Longrightarrow\ \ \textsf{LHS}$

$\huge\textsc{\underline{\underline{Hence Proved!!!}}}$

Answer 2

Answer:

Step-by-step explanation: