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Maths-

General

Easy

Question

A library has a copies of one book, b copies of each of two books, c copies of each of three books, and single copies of d books. The total number of ways in which these books can be distributed is-

$\frac{(a + b + c + d)!}{a! b! c!}$
$\frac{(a + 2 b + 3 c + d)!}{a! (b!)^{2} (c!)^{3}}$
$\frac{(a + 2 b + 3 c + d)!}{a! b! c!}$
none of these

The correct answer is: $fraction numerator left parenthesis a plus 2 b plus 3 c plus d right parenthesis factorial over denominator a factorial left parenthesis b factorial right parenthesis to the power of 2 end exponent left parenthesis c factorial right parenthesis to the power of 3 end exponent end fraction$

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