The problem:
If n, m equally spaced straight lines, mutually perpendicular to each other, bounds a rectangle, in which another rectangle is chosen and shaded with a minimum side length of $[\frac{n}{2}], [\frac{m}{2}]$. Then:
(a) How many rectangles can be chosen in the larger rectangle so that none of them coincide / contain the shaded rectangle?
(b) Probability of choosing a rectangle which contains both the shaded rectangle and at least one unshaded rectangle?
([.] - is the greatest integer function)
These are my efforts on the problem:
- I found out that the bigger rectangle has $\frac{n(n-1)m(m-1)}{4}$ possible rectangles to choose from.
- Taking a simpler case in which the inner rectangle, fixed at one of the vertices with the smaller rectangle bounded by i, j lines I used the above formula to get $$\frac{m(m-1)(n-i)(n-i-1) + n(n+1)(m-j-1)}{4} - \frac{(n-i)(n-i-1)(m-j)(m-j-1)}{4}$$
(Correct me if I’m wrong)
But, how do i find all possible ways? This is the point where i was stumped.
Please suggest a method to solve this question using only high school level combinatorics, any help would be appreciated.
(Note: This is a slightly generalised question asked in the real problem so there may be some mistakes and the title is based on what I felt!)
As you have already found out, the total number of possible shaded rectangles is $$ \frac{n(n-1)m(m-1)}{4}=\binom{n}{2}\binom{m}{2} $$ To see this, note that for the boundaries of a possible shaded rectangle, we can independantly choose $2$ of the one group of $n$ parallel straight lines (let's denote them as "vertical") and $2$ of the other (perpendicular) group of $m$ parallel lines (let's call them "horizontal").
A specific shaded rectangle is uniquely defined by its "distances" from the outermost vertical and horizontal lines. Let's denote these with $l$ and $r$ ("left" and "right") for the vertical and $t$ and $b$ ("top" and "bottom") for the horizontal boundaries. Clearly, if the selected rectangle is non-degenerated, $l+r<n-1$ and $t+b<m-1$.
First, let's count the number of "outer" rectangles that coincide/contain our specific shaded rectangle. For such an "outer" rectangle, we have $l+1$ choices for the left boundary, $r+1$ choices for the right boundary, $t+1$ choices for the top boundary and $b+1$ choices for the bottom boundary. Therefore, the number of "outer" rectangles is $(l+1)(r+1)(t+1)(b+1)$, and so the number of possible rectangles that do not coincide/contain our shaded rectangle is $$ \binom{n}{2}\binom{m}{2}-(l+1)(r+1)(t+1)(b+1) $$ Assuming that an "unshaded" rectangle is a rectangle that is not contained in (or equals) the shaded rectangle, then all rectangles that contain/coincide the shaded rectangle, except the shaded rectangle itself, also contain an unshaded rectangle, so the desired probability is $$ p=\frac{(l+1)(r+1)(t+1)(b+1)-1}{\binom{n}{2}\binom{m}{2}} $$