A population of $a$ indistinguishable alphas and $b$ indistinguishable betas can be arranged in $\binom{a+b}{a}=\binom{a+b}{b}$ distingishable orders.
Any permutations among the alphas, or among the betas , will leave the outer appearance unchanged so that $a!b!$ permutations have the same outer appearance.
It follows that if we attribute to each of the $(a+b)!$ permutations the same probability $\dfrac1{(a+b)!}$, then all distinguishable arrangements are equally probable, each having probability $\dfrac{a!b!}{(a+b)!}$.
Thus, if we speak about equally probable arrangements, the term applies both to distinguishable arrangements and to the aggregate of all permutations of the elements.
I didn't understand what is the probability $\dfrac{a!b!}{(a+b)!}$ means. What does the meaning of last sentence of the above paragraph?
The probability of a distinguishable order is the sum of the probabilties of all permutations that look the same. Since every distinguishable order has the same number of permutations that look the same, $a!b!$, the probability of every distinguishable order is the same,$$a!b!\frac{1}{(a+b)!}={{a!b!}\over{(a+b)!}}=\frac{1}{\binom{a+b}{a}}$$. The last paragraph states that when someone speaks about equally probable arrangements, they could mean the distinguishable arrangements or all permutations, from which some are indistinguishable to each other.