A) 6
B) 2
C) 3
D) 4
E) 5
Correct Answer: C
Solution :
\[^{n}{{C}_{r-1}}=28{{,}^{n}}{{C}_{r}}={{56.}^{n}}{{C}_{r+1}}=70\] \[\Rightarrow \] \[\frac{^{n}{{C}_{r}}}{^{n}{{C}_{r-1}}}=\frac{56}{28}\] \[\Rightarrow \] \[\frac{\frac{n!}{r(r-1)!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)(n-r)!}}=2\] \[=n-r+1=2r\] \[\Rightarrow \] \[n-3r=-1\] ?.. (i) \[\frac{^{n}{{C}_{r+1}}}{^{n}{{C}_{r}}}=\frac{70}{56}\] \[\Rightarrow \] \[\frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{n!}{r!(n-r)!}}=\frac{35}{28}\] \[\Rightarrow \] \[\frac{\frac{n!}{(r+1)r!(n-r-1)!}}{\frac{n!}{r!(n-r)(n-r-1)!}}=\frac{35}{28}\] \[\Rightarrow \] \[\frac{n-r}{r+1}=\frac{35}{28}\] \[\Rightarrow \] \[28n-28r=35r+35\] \[\Rightarrow \] \[28n-63r=35\] ...(ii) Multiply by 21 in Eq. (i) and subtracting from Eq. (ii), \[\begin{align} & \underline{\begin{align} & 21n-63r=-21 \\ & 28n-63r=35 \\ & -\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\, \\ \end{align}} \\ & -7n=-56\Rightarrow n=8 \\ \end{align}\] From Eq. (i), \[3r=n+1=8+1\] \[r=3\]
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