X and Y

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]

Brother OA is D.

Can you please Tell me How to Solve
x^2C4

and i am posting a link I will be very pleased if you can solve this one. I already have one expert reply on that but still not able to understand it.

https://www.beatthegmat.com/24600-t66943.html#301213

goyalsau wrote:Brother OA is D.

Can you please Tell me How to Solve
x^2C4

and i am posting a link I will be very pleased if you can solve this one. I already have one expert reply on that but still not able to understand it.

https://www.beatthegmat.com/24600-t66943.html#301213

If nC4 = n! / [4! (n – 4)!]

= n (n – 1) (n – 2) (n – 3) (n – 4)! / [4! (n – 4)!]

= n (n – 1) (n – 2) (n – 3) /24,

then x^2C4 = x^2 (x^2 – 1) (x^2 – 2) (x^2 – 3) /24

sanju09 wrote:

goyalsau wrote:Brother OA is D.

Can you please Tell me How to Solve
x^2C4

and i am posting a link I will be very pleased if you can solve this one. I already have one expert reply on that but still not able to understand it.

https://www.beatthegmat.com/24600-t66943.html#301213

If nC4 = n! / [4! (n – 4)!]

= n (n – 1) (n – 2) (n – 3) (n – 4)! / [4! (n – 4)!]

= n (n – 1) (n – 2) (n – 3) /24,

then x^2C4 = x^2 (x^2 – 1) (x^2 – 2) (x^2 – 3) /24

Excellent, what is healthier about you is that you are happy helping persons with lack of gratitude as well. Keep the good work up sanju09