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

A

$15$ અથવા $8$

B

$10$ અથવા $5$

C

$15$ અથવા $10$

D

$8$ અથવા $10$

Solution

$\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  4 
\end{array}} \right)\,\, + \,\,\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  6 
\end{array}} \right)\,\, = \,\,2\,\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  5 
\end{array}} \right)\,$

$\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  4 
\end{array}} \right)\,\, + \,\,\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  5 
\end{array}} \right)\,\, + \,\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  5 
\end{array}} \right)\,\,\, + \,\,\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  6 
\end{array}} \right)\,\,\, = \,\,4\,\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  5 
\end{array}} \right)\,\,\,\,\,\,$

$\left( {_5^n} \right)\, + \,\,\left( {_6^n} \right)\,\, = \,\,4\,\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  5 
\end{array}} \right)\,$

 

$\,\left( {\begin{array}{*{20}{c}}
  {n + 1} \\ 
  6 
\end{array}} \right)\,\,\, = \,4\,\left( {\begin{array}{*{20}{c}}
  {n – 1} \\ 
  5 
\end{array}} \right)\,\,\,\,$

$\frac{{n\,(n + 1)(n – 1)(n – 2)(n – 3)(n – 4)}}{{720}}$

$\, = \,\,\frac{{4(n – 1)(n – 2)(n – 3)(n – 4)(n – 5)}}{{120}}$

$n^2 + n = 24(n – 5) $

$n^2 – 23n + 120 = 0$ 

$ n = 15 $ અથવા $ 8$

 

Standard 11
Mathematics

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