2nPn+1:2n-2Pn=56:3 find n
Answers
Answer:
By solving this equation \frac{(2 n+1) P_{n-1}}{(2 n-1) P_{n}}=\frac{3}{5}
(2n−1)P
n
(2n+1)P
n−1
=
5
3
, 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 NnP
r
=
(n−r)!
n!
, where ,1≤r≤n,n∈N
We have,
\frac{(2 \mathrm{n}+1) \mathrm{P}_{\mathrm{n}-1}}{(2 \mathrm{n}-1) \mathrm{P}_{\mathrm{n}}}=\frac{3}{5}
(2n−1)P
n
(2n+1)P
n−1
=
5
3
\Rightarrow \frac{(2 \mathrm{n}+1) \mathrm{p}_{\mathrm{n}-1}}{3}=\frac{(2 \mathrm{n}-1) \mathrm{P} \mathrm{n}}{5}⇒
3
(2n+1)p
n−1
=
5
(2n−1)Pn
\Rightarrow \frac{(2 n+1) !}{((2 n+1)-(n-1)) ! \times 3}=\frac{(2 n-1) !}{((2 n-1)-n) ! \times 5}⇒
((2n+1)−(n−1))!×3
(2n+1)!
=
((2n−1)−n)!×5
(2n−1)!
\Rightarrow \frac{(2 n+1)(2 n)(2 n-1) !}{(n+2) ! \times 3}=\frac{(2 n-1) !}{(n-1) ! \times 5}⇒
(n+2)!×3
(2n+1)(2n)(2n−1)!
=
(n−1)!×5
(2n−1)!
\Rightarrow \frac{(2 n+1)(2 n)}{(n+2)(n+1) n((n-1) !) \times 3}=\frac{1}{(n-1) ! \times 5}⇒
(n+2)(n+1)n((n−1)!)×3
(2n+1)(2n)
=
(n−1)!×5
1
\Rightarrow \frac{4 n+2}{(n 2+3 n+2) \times 3}=\frac{1}{5}⇒
(n2+3n+2)×3
4n+2
=
5
1
\begin{gathered}\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}\end{gathered}
⇒20n+10=3n
2
+9n+6
⇒3n
2
−11n−4=0
⇒3n
2
−12n+n−4=0
\begin{gathered}\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}\end{gathered}
⇒3n(n−4)+1(n−4)=0
⇒(n−4)(3n+1)=0
⇒n−4=0 or 3n+1=0
⇒n=4 or n=−13∈
/
N
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.
Step-by-step explanation:
hope it helps...
Answer:
final answer =4
Step-by-step explanation:
refer to photo
