goyalsau wrote:Hi! guys.
Great finding, Goyalstau...
The minimum possible number of bulbs in the given square grid with bulbs as fixed is therefore 9, but we need to deal in x only.
The question is about selecting the particular 4 bulbs out of many selections of 4 out of a given x^2 bulbs, where x is an integer greater than 2. The total ways of selecting 4 bulbs on the grid is
= x^2C4
= x^2 (x^2 - 1) (x^2 - 2) (x^2 - 3) /24
Now, try taking clues in terms of x while keep asking you of how many 2 × 2 square grids may be seen in your mind's eye on a 3 × 3 square grid, on a 4 × 4 square grid, on a 5 × 5 square grid, and so on... and hence on an x × x square grid.
Let's give it a try
On a 3 × 3 square grid, I can dream 4 sets of a 2 × 2 square grid. This simply happens to be the square of the number of allotments allowed to a 2 × 2 square grid along any edge of an x × x square grid.
On any edge of a 3 × 3 square grid, a 2 × 2 square grid can have only 2 allotments, whose square is the number of 2 × 2 square grids that may be dreamt.
It is evident that on an x × x square grid, x - 1 such allotments are possible, hence (x - 1) ^2 number of 2 × 2 square grids are possibly seen on it. Favorable number of ways
= (x - 1) ^2
The required probability
= 24 (x - 1) ^2/ x^2C4
= 24 (x - 1) ^2/ [x^2C4 = x^2 (x^2 - 1) (x^2 - 2) (x^2 - 3)]
= [spoiler]
24 (x - 1)/ [x^2 (x + 1) (x^2 - 2) (x^2 - 3)]
= which choice should I go for, is something blurred...
Search it in answers somebody, please[/spoiler]
Last edited by
sanju09 on Sat Sep 25, 2010 5:44 am, edited 1 time in total.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
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