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If a set $A$ has $n$ elements in it, how many reflexive relations can be defined on it?

My solution

Is the answer

summation of (n^2 - n)C(i) for i=0 to n^2 -n

$$\sum_{i=0}^{n^2-n} C(n^2-n,i) = \sum_{i=0}^{n^2-n} \binom{n^2-n}i$$

well if i make a matrix $n\times n$ now the diagonal elements have to be selected,out of remaining $n^2-n$ any number of elements can be selected.

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Reasoning is OK. There is a much simpler-looking answer based on the same idea. Or else you can simplify your expression and then notice that it is easier than it looks. –  André Nicolas Dec 3 '11 at 15:33
@Andre Nicolas that would be? –  Kraken Dec 3 '11 at 15:35
@Karan: I've tried to change your question to use TeX for better readability, but I left also your original text for the equation with the suggested answer. Please, check whether my edits did not change the meaning of your question and, if necessary, edit your question. –  Martin Sleziak Dec 3 '11 at 15:39
the answer is correct, right?? –  Kraken Dec 3 '11 at 15:39
These questions are similar to this one: math.stackexchange.com/questions/12139/… and math.stackexchange.com/questions/15108/… –  Martin Sleziak Dec 3 '11 at 15:40
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1 Answer

up vote 2 down vote accepted

Your idea is fine. For each place $(i,j)$ in the matrix that is not on the diagonal, say YES or NO depending on whether you want $R(i,j)$ to hold or not to hold. There are $2^{n^2-n}$ different ways of doing this.

Note that one can get that answer from yours. For in general, by the Binomial Theorem, $$(1+x)^m=\sum_{i=0}^m \binom{m}{i}x^i.$$ Put $m=n^2-n$ and $x=1$. On the right we get your expression, and on the left we get $2^{n^2-n}$.

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yeah.. thanks.. :) –  Kraken Dec 3 '11 at 15:41
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