If \[{}^{n}{{C}_{r}}={}^{n}{{C}_{r-1}}\]and \[{}^{n}{{P}_{r}}={}^{n}{{P}_{r+1}},\]then the value of \[n\] is |
A) 3
B) 4
C) 2
D) 5
Correct Answer: A
Solution :
\[{}^{n}{{C}_{r}}={}^{n}{{C}_{r-1}}\] |
\[\Rightarrow \]\[\frac{n!}{(n-r)!r!}=\frac{n!}{(n-r+1)!(r-1)!}\] |
\[\Rightarrow \] \[1=\frac{(n-r)!r\,\,(r-1)!}{(n-r+1)(n-r)!(r-1)!}=\frac{r}{(n-r+1)}\] |
\[\Rightarrow \]\[n-r+1=r\] (i) |
Again, \[{}^{n}{{P}_{r}}={}^{n}{{P}_{r+1}}\] |
\[\Rightarrow \]\[\frac{n!}{(n-r)!}=\frac{n!}{(n-r-1)!}\] |
\[\Rightarrow \]\[\frac{1}{(n-r)(n-r-1)!}=\frac{1}{(n-r-1)!}\] |
\[\Rightarrow \] \[n-r=1\] ... (ii) |
From Eqs. (i) and (ii), we get |
\[r=2\] |
\[\therefore \] \[n=3\] |
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