A) 10
B) 9
C) 19
D) None
Correct Answer: B
Solution :
\[\frac{{{P}_{n+1}}}{{{P}_{n}}}=\frac{^{n+1}{{C}_{0}}^{n+1}{{C}_{1}}^{n+1}{{C}_{2}}^{n+1}...{{C}_{n+1}}}{^{n}{{C}_{0}}{{.}^{n}}{{C}_{1}}{{.}^{n}}{{C}_{2}}{{...}^{n}}{{C}_{n}}}\] \[=\frac{^{n+1}{{C}_{1}}}{^{n}{{C}_{1}}}.\frac{^{n+1}{{C}_{2}}}{^{n}{{C}_{2}}}...\frac{^{n+1}{{C}_{2}}}{^{n}{{C}_{n}}}{{.}^{n+1}}{{C}_{n-1}}\] \[=\frac{n+1}{n}.\frac{n+1}{n-1}.\frac{n+1}{n-2}........\frac{n+1}{n}=\frac{{{\left( n+1 \right)}^{n}}}{n!}\] Given\[\frac{{{P}_{n+1}}}{{{P}_{n}}}=\frac{{{10}^{9}}}{9!}\Rightarrow \frac{{{(n+1)}^{n}}}{n!}=\frac{{{10}^{9}}}{9!}\]\[\therefore \]\[n=9\]
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