If ^n{C_4},{,^n}{C_5}, and {,^n}{C_6},&n | vedclass

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Question

$2.{\,^n}{C_5}{ = ^n}{C_4}{ + ^n}{C_6}$

$2.\frac{{\left| n \right.}}{{\left| {5\left| {n - 5} \right.} \right.}} = \frac{{\left| n \right.}}{{\left

Vedclass · Accepted Answer

$14$

Vedclass · Answer

$2.{\,^n}{C_5}{ = ^n}{C_4}{ + ^n}{C_6}$

$2.\frac{{\left| n \right.}}{{\left| {5\left| {n - 5} \right.} \right.}} = \frac{{\left| n \right.}}{{\left| {4\left| {n - 4} \right.} \right.}} + \frac{{\left| n \right.}}{{\left| {6\left| {n - 6} \right.} \right.}}$

$\frac{2}{5}.\frac{1}{{n - 5}} = \frac{1}{{\left( {n - 4} \right)\left( {n - 5} \right)}} + \frac{1}{{30}}$

$n=14$ satisfying equation.

If $^n{C_4},{\,^n}{C_5},$ and ${\,^n}{C_6},$ are in $A.P.,$ then $n$ can be

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