Light bulbs

IMO B

This is a VIC problem, so you can solve it by picking numbers or by algebra. Normally when the algebra is difficult the best approach is to pick numbers.

Let's pick x = 4. Now we have to calculate the probability that in a bulb 4x4,in which 4 bulbs are illuminated at random,these bulbs form a 2x2 bulb square.

Probability = desire outcomes/total possibilities.

total possibilities = C(16,4) = 16!/12!4! = 4 x 5 x 7 x 13

now we have to calculate how many of those combinations represent a 2x2 bulb square. If you don't know how to calculate this you can just draw the square and count. If you do so, you'll see that the combinations are 9. In fact the formula is (x-1)^2

the probability is = 9/ (4x5x7x13)

now we have to plug 4 in every of the solutions

if you plug it in B

24 (4-1) / (4^2)(4^2-2)(4^2-3)(4 +1) = 72/(16 * 14 * 13 * 5) = 9 /(4*7*13*5)


You can solve the problem using algebra and then you have to calculate everything based on X

total combinations = C(X^2,4)= X^2!/(X^2-4)!4! =
(X^2) * (X^2-1) * (X ^2 - 2) * (X ^2-3)/ 4!
you can notice that (X^2-1) = (X-1)(X+1)

desire outcomes = (X-1)(X-1)

Probability = (X-1)(X-1)/(X^2)(X-1)(X+1) (X ^2 - 2) * (X ^2-3)/ 4! = 4!(X-1)/(X^2)(X+1) (X ^2 - 2) (X ^2-3)

now we have to calculate how many of those combinations represent a 2x2 bulb square.

You said formula is (x-1)^2. How did u get this formula?? Also what if i wanted to find out combinations of 3x3. How to do this?

muna_m wrote:now we have to calculate how many of those combinations represent a 2x2 bulb square.

You said formula is (x-1)^2. How did u get this formula?? Also what if i wanted to find out combinations of 3x3. How to do this?

How many 1*1 (i.e. how many items in the grid) is x^2
How many 2*2 is (x-1)^2
How many 3*3 is (x-2)^2

and so on, and so on...

Oh yea! Thanks