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STD 11 - 6. permutation and combination — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 6. permutation and combination1 Mark
MCQ
There is a rectangular sheet of dimension $(2m - 1) × (2n - 1)$, (where $m > 0,n > 0)$. 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
  • A
    ${(m + n + 1)^2}$
  • B
    $mn(m + 1)\,(n + 1)$
  • C
    ${4^{m + n - 2}}$
  • ✓
    ${m^2}{n^2}$
✓

Answer

Correct option: D.
${m^2}{n^2}$
d
(d) Along horizontal side one unit can be taken in $(2m-1)$ ways and $3$ unit side can be taken in $2m - 3$ ways.

The number of ways of selecting a side horizontally is

$(2m - 1 + 2m - 3 + 2m - 5 + .... + 3 + 1)$

Similarly the number of ways along vertical side is $(2n - 1 + 2n - 3 + .... + 5 + 3 + 1)$.

$\therefore$ Total number of rectangles

$ = [1 + 3 + 5 + ..... + (2m - 1)] \times [1 + 3 + 5 + .... + (2n - 1)]$

$ = \frac{{m(1 + 2m - 1)}}{2} \times \frac{{n(1 + 2n - 1)}}{2} = {m^2}{n^2}$.

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