A) 3
B) 2
C) 1
D) 0
E) 4
Correct Answer: B
Solution :
\[\frac{1}{^{4}{{C}_{n}}}=\frac{1}{^{5}{{C}_{n}}}+\frac{1}{^{6}{{C}_{n}}}\] \[\Rightarrow \]\[\frac{n!(4-n)!}{4!}=\frac{n!(5-n)!}{5!}+\frac{n!(6-n)!}{6!}\] \[\Rightarrow \] \[\frac{(4-n)!}{4!}=\frac{(4-n)!(5-n)}{5\times 4!}\] \[+\frac{(6-n)(5-n)(4-n)!}{6\times 5\times 4!}\] \[\Rightarrow \] \[1=\frac{5-n}{5}+\frac{(6-n)(5-n)}{6\times 5}\] \[\Rightarrow \] \[30=30-6n+30-11n+{{n}^{2}}\] \[\Rightarrow \] \[{{n}^{2}}-17n+30=0\] \[\Rightarrow \] \[(n-15)(n-2)=0\] \[\Rightarrow \] \[n=2\] (\[\because \]\[^{4}{{C}_{n}}\]is not meaningful for\[n=15\])
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