the number of different nxn symmetric matrices with each element being either 0 or 1 is ______
a. 2^n
b. 2^(n^2)
c.2^((n^2+n)/2)
d.2^((n^2-n)/2) @IIT study group

Question

the number of different nxn symmetric matrices with each element being either 0 or 1 is ______
a. 2^n
b. 2^(n^2)
c.2^((n^2+n)/2)
d.2^((n^2-n)/2)    @IIT study group

cathyangs · Answer

I'm pretty sure that it's b.

anonymous · Answer

how?

anonymous · Answer

Option C.

$$ \large \huge 2 ^ {\frac{(n^2 + n )}{2}} $$

anonymous · Answer

yes... c is the correct answer... how would u say that? please tell me...

anonymous · Answer

I know .. ;)

cathyangs · Answer

for each element, it can be either 0 or 1. Because there are nxn elements in the matrix, AHH!! it's symmetric. sorry, its 2^((n^2+n)/2)

anonymous · Answer

thank you... :)

anonymous · Answer

but can u elaborate the reason please...

anonymous · Answer

Alright .. how many places in the matrix can have 0 or 1?

anonymous · Answer

but can u elaborate the reason please...

anonymous · Answer

yaa... got it... thanks... it is (n^2+n)/2 places in which either 1 or 0 can be present... the remaining elements depends on the filled in elements as the matrix is symmetric... so... c is the option...

anonymous · Answer

anusha?

anonymous · Answer

but can u elaborate the reason please...

anonymous · Answer

thank u @ FoolForMath and @cathyangs

anonymous · Answer

You are welcome :)