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Maths-

General

Easy

Question

The streets of a city are arranged like the lines of a chess board. There are m streets running North to South & 'n' streets running East to West. The number of ways in which a man can travel from NW to SE corner going the shortest possible distance is-

$\sqrt{m^{2} + n^{2}}$
$\sqrt{(m - 1)^{2} . (n - 1)^{2}}$
$\frac{(m + n)!}{m! . n!}$
$\frac{(m + n - 2)!}{(m - 1)! . (n - 1)!}$

The correct answer is: $fraction numerator left parenthesis m plus n – 2 right parenthesis factorial over denominator left parenthesis m – 1 right parenthesis factorial. left parenthesis n – 1 right parenthesis factorial end fraction$

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