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6.Permutation and Combination
medium
यदि $P(n,r) = 1680$ और $C(n,r) = 70$, तब $69n + r! = $
A
$128$
B
$576$
C
$256$
D
$625$
Solution
$P(n,r) = 1680$ $\frac{{n!}}{{(n – r)!}} = 1680$…..$(i)$
$C(n,r) = 70$ $\Rightarrow\, \frac{{n!}}{{r!\,(\,n – r)!}} = 70$…..$(ii)$ $\frac{{1680}}{{r!}} = 70$, [$(i)$ व $(ii)$ से]
$r! = \frac{{1680}}{{70}} = 24$ $⇒ r = 4$
$\because $ $p(n,4) = \,1680$
$\therefore $ $n(n – 1)(n – 2)(n – 3) = 1680$ $\Rightarrow$ $n = 8$
अत: $8 \times 7 \times 6 \times 5 = 1680$
इस प्रकार, $69n + r\,! = 69 \times 8 + 4!$
$ = 552 + 24$ $= 576$
Standard 11
Mathematics