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જો $\left( {_3^n} \right) + \left( {_4^n} \right) > \left( {_{\,\,\,3}^{n + 1}} \right)$ હોય, તો....
$n > 6$
$n > 7$
$n < 6$
$n < 5$
Solution
$\left( {\begin{array}{*{20}{c}}
n \\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
n \\
4
\end{array}} \right) > \left( {\begin{array}{*{20}{c}}
{n + 1} \\
3
\end{array}} \right)$
$\left( {\begin{array}{*{20}{c}}
{n + 1} \\
4
\end{array}} \right) > \left( {\begin{array}{*{20}{c}}
{n + 1} \\
3
\end{array}} \right)$
$\frac{{(n + 1)\,!}}{{4\,!\,(n – 3)\,!}} > \,\frac{{(n + 1)\,!}}{{3\,!\,(n – 2)\,!}}\,\,$
$\,\,\frac{{(n + 1)\,!}}{{4\,(3\,!)(n – 3)\,!}} > \,\frac{{(n + 1)\,!}}{{3\,!\,(n – 2)(n – 3)\,!}}\,\,\,$
$\,\,\frac{1}{4} > \frac{1}{{n – 2}}\,\,$
$4 < n – 2$
$6 < n$
$\,\,n > 6$
