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6.Permutation and Combination
easy
यदि $^{2n}{C_3}:{\,^n}{C_2} = 44:3$ हो, तो $r$ के किस मान के लिये $^n{C_r}$ का मान 15 होगा
A
$r = 3$
B
$r = 4$
C
$r = 6$
D
$r = 5$
Solution
$\frac{{(2n)\;!}}{{(2n – 3)\;!\;.\;3\;!}} \times \frac{{2\;!\; \times (n – 2)\;!}}{{n\;!}}\; = \frac{{44}}{3}$ $ \Rightarrow $ $n$
$\Rightarrow 4(2n – 1) = 44$ $ \Rightarrow $ $2n = 12$ $ \Rightarrow $$n = 6$
अब $^6{C_r} = 15$ $ \Rightarrow $ $^6{C_r}{ = ^6}{C_2}$ या $^6{C_4}$$ \Rightarrow $ $r = 2,\;4$.
Standard 11
Mathematics
