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6.Permutation and Combination
hard
$^{n - 1}{C_r} = ({k^2} - 3)\,.{\,^n}{C_{r + 1}}$, यदि $k \in $
A
$[ - \sqrt 3 ,\,\sqrt 3 ]$
B
$( - \infty ,\, - 2)$
C
$(2,\,\infty )$
D
$(\sqrt 3 ,\,2)$
(IIT-2004)
Solution
$\frac{{(n – 1)\,!}}{{(n – r – 1)\,!\,r\,!}} = \frac{{({k^2} – 3)\,n\,!}}{{(n – r – 1)\,!\,(r + 1)\,!}}$, $0 \le r \le n – 1$
$ \Rightarrow $${k^2} = \frac{{r + 1}}{n} + 3,\,\frac{1}{n} \le \frac{{r + 1}}{n} \le 1$
==> ${k^2} \in \left[ {\frac{1}{n} + 3,\,4} \right]\,,\,n \ge 2$
$k \in \left[ { – 2,\, – \sqrt {\frac{1}{n} + 3} } \right] \cup \left[ {\sqrt {\frac{1}{n} + 3} ,\,2} \right];\,n \ge 2$.
Standard 11
Mathematics
