The numbers $1, 2, 3, 4, . . . . . . n$ are arranged in a row at random. The probability that the digits $1, 2, 3, . . . . . . k (k < n)$ appear as neighbours in that order is

\frac{(n - k + 1)!}{n!}

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\frac{(n - k + 1)!}{(n - k)!}

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\frac{(n - k + 1)! k!}{n!}

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\frac{(n - k + 1)! k!}{(n - k)!}

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Solution

The correct option is A $\frac{(n - k + 1)!}{n!}$
$T o t a l n u m b e r o f a r r a n g e m e n t s o f n n u m b e r s = n! I f w e w a n t^{'} k^{'} n u m b e r s t o a p p e a r i n t h e s p e c i f i c o r d e r i . e . 1, 2, 3, . . . k, t h e n i t f o r m s o n e o b j e c t a n d r e s t o f (n - k) n u m b e r s c a n b e a r r a n g e d i n a n y o r d e r . ∴ R e q d . n o . o f a r r a n g e m e n t s = (n - k + 1)!$ $(h e r e 1 a d d e d ∵ (n - k) n u m b e r s a n d 1 o b j e c t o f 1 t o k n u m b e r s w i l l b e a r r a n g e d) ∴ P r o b a b i l i t y = \frac{(n - k + 1)!}{n!} ∴ O p t i o n (A) i s t h e a n s w e r .$

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