1.

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

Answer»

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!!!}}}


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