For each integer i, $1 \leq i \leq 100 ; \varepsilon_{i}$ be either +1 or $-1 .$ Assume that $\varepsilon_{1}=+1$ and $\varepsilon_{100}=-1 .$ Say that a sign change occurs at $i \geq 2$ if $\varepsilon_{i}, \varepsilon_{i-1}$ are of opposite sign. Then the total number of sign changes
(a) is odd
(b) is even
(c) is at most 50
(d) can have 49 distinct values
My logic
$+$ to $+$ sign change is even, as + to – and then – to +, now, $ԑ_{100}$ is -1. So, number of sign changes must be odd.
Yes, suppose you perform sign change $m$ times.
Then $$-1=\epsilon_{100} = (-1)^m \epsilon_1=(-1)^m$$
$m$ has to be odd.
For example, we can have sign change at every step that is $\epsilon_{2i+1} = 1$ and $\epsilon_{2i}=-1$, in that case we have $99$ sign change.
At the other extreme end, we can have exactly one sign change, say $\epsilon_1=-1$ and $\epsilon_i=1, i \ge 2$, in that case, we can have $1$ sign change.
We can have $\{1,3,5,\ldots, 99\}$, I will leave the computation of the cardinality of the set for you to disprove $(d)$.