Question

find n if 2n+1 P n-1 : 2n-1Pn = 3:5

Answer 1

Please check the attachments. Thank you.

Answer 2

By solving this equation $\frac{(2 n+1) P_{n-1}}{(2 n-1) P_{n}}=\frac{3}{5}$, the ‘value’ of n is “4”.  

$\mathrm{n} \mathrm{P}_{\mathrm{r}}=\frac{n !}{(n-r) !}, \text { where }, 1 \leq \mathrm{r} \leq \mathrm{n}, n \in N$

We have,

$\frac{(2 \mathrm{n}+1) \mathrm{P}_{\mathrm{n}-1}}{(2 \mathrm{n}-1) \mathrm{P}_{\mathrm{n}}}=\frac{3}{5}$

$\Rightarrow \frac{(2 \mathrm{n}+1) \mathrm{p}_{\mathrm{n}-1}}{3}=\frac{(2 \mathrm{n}-1) \mathrm{P} \mathrm{n}}{5}$

$\Rightarrow \frac{(2 n+1) !}{((2 n+1)-(n-1)) ! \times 3}=\frac{(2 n-1) !}{((2 n-1)-n) ! \times 5}$

$\Rightarrow \frac{(2 n+1)(2 n)(2 n-1) !}{(n+2) ! \times 3}=\frac{(2 n-1) !}{(n-1) ! \times 5}$

$\Rightarrow \frac{(2 n+1)(2 n)}{(n+2)(n+1) n((n-1) !) \times 3}=\frac{1}{(n-1) ! \times 5}$

$\Rightarrow \frac{4 n+2}{(n 2+3 n+2) \times 3}=\frac{1}{5}$

$\begin{array}{l}{\Rightarrow 20 n+10=3 n^{2}+9 n+6} \\ {\Rightarrow 3 n^{2}-11 n-4=0} \\ {\Rightarrow 3 n^{2}-12 n+n-4=0}\end{array}$

$\begin{array}{l}{\Rightarrow 3 n(n-4)+1(n-4)=0} \\ {\Rightarrow(n-4)(3 n+1)=0} \\ {\Rightarrow n-4=0 \text { or } 3 n+1=0} \\ {\Rightarrow n=4 \text { or } n=-13 \notin N}\end{array}$

So,

n = 4

Permutation is defined as arranging the ‘elements of a set’ into a ‘sequence’ or ‘order’. Permutations are different from combinations, which are “selections of some elements” of a set regardless of order.