A) \[8\]
B) \[9\]
C) \[10\]
D) \[11\]
Correct Answer: B
Solution :
Given that, \[^{n}{{C}_{r-1}}=36,\,\,\,\,{{\,}^{n}}{{C}_{r}}=84\] and \[^{n}{{C}_{r+1}}=126\] \[\Rightarrow \] \[\frac{n!}{(n-r+1)!}=36\], \[\frac{n!}{(n-r)!r!}=84\] and \[\frac{n!}{(n-r-1!)(r+1)!}=126\] Now, \[\frac{(n-r)!r!}{(n-r+1)!(r-1)!}=\frac{36}{84}\] \[\Rightarrow \] \[\frac{r}{(n-r+1)}=\frac{3}{7}\] \[\Rightarrow \] \[10r-3n-3=0\] ? (i) and \[\frac{(n-r-1)!(r+1)!}{(n-r)!r!}=\frac{84}{126}\] \[\Rightarrow \] \[\frac{r+1}{(n-r)}=\frac{2}{3}\] \[\Rightarrow \] \[5r-2n+3=0\] ? (ii) On solving Eqs. (i) and (ii), we get \[r=3,\] \[n=9\]
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