Maths-
General
Easy

Question

There is a rectangular sheet of dimension left parenthesis 2 m minus 1 right parenthesis ×left parenthesis 2 n minus 1 right parenthesis, (where m greater than 0 comma n greater than 0 right parenthesis. It has been divided into square of unit area by drawing lines perpendicular to the sides. Find number of rectangles having sides of odd unit length

  1. left parenthesis m plus n plus 1 right parenthesis to the power of 2 end exponent    
  2. m n left parenthesis m plus 1 right parenthesis left parenthesis n plus 1 right parenthesis    
  3. 4 to the power of m plus n minus 2 end exponent    
  4. m to the power of 2 end exponent n to the power of 2 end exponent    

The correct answer is: m to the power of 2 end exponent n to the power of 2 end exponent


    Along horizontal side one unit can be taken in (2m–1) ways and 3 unit side can be taken in 2 m minus 3 ways.
    \ The number of ways of selecting a side horizontally is
    left parenthesis 2 m minus 1 plus 2 m minus 3 plus 2 m minus 5 plus.... plus 3 plus 1 right parenthesis
    Similarly the number of ways along vertical side is left parenthesis 2 n minus 1 plus 2 n minus 3 plus.... plus 5 plus 3 plus 1 right parenthesis.

    \Total number of rectangles
    equals left square bracket 1 plus 3 plus 5 plus..... plus left parenthesis 2 m minus 1 right parenthesis right square bracket cross times left square bracket 1 plus 3 plus 5 plus.... plus left parenthesis 2 n minus 1 right parenthesis right square bracket
    equals fraction numerator m left parenthesis 1 plus 2 m minus 1 right parenthesis over denominator 2 end fraction cross times fraction numerator n left parenthesis 1 plus 2 n minus 1 right parenthesis over denominator 2 end fraction equals m to the power of 2 end exponent n to the power of 2 end exponent.

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