જો binom{n-1}{4} , binom{n-1}{5} ,binom{ | vedclass

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Question

$\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  4 
\end{array}} ight)\,\, + \,\,\left( {\begin{array}{*{20}{c}}
  {n -

Vedclass · Accepted Answer

$15$ અથવા $8$

Vedclass · Answer

$\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  4 
\end{array}} ight)\,\, + \,\,\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  6 
\end{array}} ight)\,\, = \,\,2\,\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  5 
\end{array}} ight)\,$

$\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  4 
\end{array}} ight)\,\, + \,\,\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  5 
\end{array}} ight)\,\, + \,\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  5 
\end{array}} ight)\,\,\, + \,\,\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  6 
\end{array}} ight)\,\,\, = \,\,4\,\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  5 
\end{array}} ight)\,\,\,\,\,\,$

$\left( {_5^n} ight)\, + \,\,\left( {_6^n} ight)\,\, = \,\,4\,\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  5 
\end{array}} ight)\,$

 

$\,\left( {\begin{array}{*{20}{c}}
  {n + 1} \ 
  6 
\end{array}} ight)\,\,\, = \,4\,\left( {\begin{array}{*{20}{c}}
  {n - 1} \ 
  5 
\end{array}} ight)\,\,\,\,$

$\frac{{n\,(n + 1)(n - 1)(n - 2)(n - 3)(n - 4)}}{{720}}$

$\, = \,\,\frac{{4(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)}}{{120}}$

$n^2 + n = 24(n - 5) $

$n^2 - 23n + 120 = 0$ 

$ n = 15 $ અથવા $ 8$

જો $\binom{n-1}{4} , \binom{n-1}{5} ,\binom{n-1}{6}$ સમાંતર શ્રેણી હોય તો $n$ શોધો

Similar Questions

જો $^{n} C_{8}=\,^{n} C_{2}$ હોય, તો $^{n} C_{2}$ શોધો.