A) 964
B) 625
C) 227
D) 232
Correct Answer: D
Solution :
\[\frac{^{n}{{C}_{r-1}}}{^{n}{{C}_{r}}}=\frac{2}{15}\] \[\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{2}{15}\] \[\frac{r}{n-r+1}=\frac{2}{15}\] \[15r=2n-2r+2\] \[\] \[\frac{^{n}{{C}_{r}}}{^{n}{{C}_{r+1}}}=\frac{15}{70}\] \[\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}}=\frac{3}{14}\] \[\frac{r+1}{n-r}=\frac{3}{14}\] \[14r+14=3n-3r\] \[\begin{align} & 3n-17r=14 \\ & \frac{2n-17r=-2}{n=16} \\ \end{align}\] \[17r=34,r=2\] \[^{16}{{C}_{1}}{{,}^{16}}{{C}_{2}}{{,}^{16}}{{C}_{3}}\] \[\frac{^{16}{{C}_{1}}{{+}^{16}}{{C}_{2}}{{+}^{16}}{{C}_{3}}}{3}=\frac{16+120+560}{3}\] \[\frac{680+16}{3}=\frac{696}{3}=232\]You need to login to perform this action.
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