# Pascal's triangle.

Is there an intuitive definition for the symmetry that occurs in Pascal's triangle? If $n$ in $\binom{n}{k}$ is odd, there is indeed an exact reflection along the center column (which divides the triangle into halves of $n-2$ parts each, and each half contains exactly the same elements in reverse order with respect to one another. This makes sense to me.

However, if $n$ is even, there is still a sort of additive? symmetry, where if you take the alternating sum of the binomial coefficients, the result is zero. This second fact is what prompted me to ask the question, as it seemed very mysterious...

Suppose n = 6, then 1 - 6 + 15 - 20 + 15 - 6 + 1 = 0, which seems very strange, as the "halves" are not broken evenly and contain no elements in common. But it may be because I'm missing something.

With regards to @Jesse Meng's answer,

Interestingly, that second formula is precisely what I was trying to understand (intuitively). I understand and can apply the formula. However, I am missing the intuition with regards to why selecting x = 1 and y = -1 signifies combinatorially the alternating sum . Is there no simple way to convey it? Is there even an intuition? Should I completely reconsider my frame?

From the binomial formula, you would have $$(1+1)^n=\sum_{k=0}^{n}\binom{n}{k}$$ Similarly, you have $$(1-1)^n=\sum_{k=0}^{n}\binom{n}{k}(-1)^k$$ Notice $n$ does not need to be even here, so you have your desired result. So, the intuition here lies within the intuition of the binomial expansion formula itself - I am certain there is a rich number of resources that can expand on the intuition of this formula.
• @RafaelVergnaud I can try to offer you some intuition from combinatorics: Suppose you have a set of n elements, then the equation becomes: the number of odd subsets$=\binom{n}{1}+\binom{n}{3}+...$ is equal to the number of even subsets $=\binom{n}{0}+\binom{n}{2}+...$. Hence, it suffices for us to understand why the number of even subsets of n = number of odd subsets of n. It turns out for each even subset, it has a corresponding "matching" odd subset. – Jesse Meng Apr 6 '18 at 4:27
• This is done so by choosing an arbitrary element from the n elements, assuming $n$ is not $0$, such an arbitrary element must exist. We fix this element and name it $x$. For either an odd/even set, we can apply a transformation:(if it has $x$, remove it, otherwise put in x) to change its size precisely by $1$, this transformation is a bijection between odd and even subsets. Hence we have that the number of odd and even subsets are equal, because every odd/even subset has its own "unique matching" Is this intuitive enough? – Jesse Meng Apr 6 '18 at 4:30
A simple explanation: you can choose $k$ objects from $n$ the same number of ways you don't choose $n-k$ from $n$ hence the symmetry.