The purpose of this kind of problem is to make you understand the formula for and the assumptions behind the binomial distribution.

The assumptions are repeated trials, only two possible outcomes of a single trial, and independence of trials.

If n is the number of trials and each trial has exactly two possible outcomes, what is the number of possible outcomes? Obviously $2^n.$

It takes nowhere near 25 minutes to build that outcome chart in your problem. You will have $2^5 = 32$ outcomes. So using g and b to indicate the different outcomes, you start as follows

g g g g g g g g g g g g g g g g b b b b b b b b b b b b b b b b$\ \ \ $ I wrote down $\dfrac{32}{2} = 16\ g's,\ then\ 16\ b's\ ONCE.$

g g g g g g g g b b b b b b b b g g g g g g g g b b b b b b b b$\ \ \ $ Below that, I wrote down $\dfrac{16}{2} = 8\ g's, then\ 8 b's\ TWICE.$

g g g g b b b b g g g g b b b b g g g g b b b b g g g g b b b b$\ \ \ $ Below that, I wrote down $\dfrac{8}{2} = 4\ g's, then\ 4 b's\ THRICE.$

g g b b g g b b g g b b g g b b g g b b g g b b g g b b g g b b$\ \ \ $ Below that, I wrote down $\dfrac{4}{2} = 2\ g's, then\ 2 b's\ FOUR\ times.$

g b g b g b g b g b g b g b g b g b g b g b g b g b g b g b g b$\ \ \ $ Below that, I wrote down $\dfrac{2}{2} = 1\ g, then\ 1 b\ EIGHT\ times.$

That is totally mechanical, no thinking required. Now reading down instead of across, I get a description of each possible outcome. The sixth column over reads downward ggbgb, meaning the first child was a girl, the second child was a girl, the third child was a boy, the fourth child was a girl, and the fifth child was a boy.

Now I can count how many g's and b's there are in each column and summarize how many columns have 5 g's, how many have 4 g's, how many have 3 g's, how many have 2 g's, how many have 1 g, and how many have no g's.

I get

1 case of 5 g's, which means 1 case of no b's.

5 cases of 4 g's, which means 5 cases of 1 b.

10 cases of 3g's, which means 10 cases of 2 b's.

10 cases of 2g's, which means 10 cases of 3 b's.

5 cases of 1 g, which means 5 cases of 4 b's.

1 case of no g's, which means 1 case of 5 b's.

If you are familiar with Pascal's triangle, you may see a pattern

Now there IS a formula for the related probabilities.

$\dbinom{n}{m} * p^m * q^{n - m}.$

So given that boys and girls have an equal probability of being born (which actually is not quite true), the probability of exactly three girls out of five births is

$\dbinom{5}{3} * \left(\dfrac{1}{2}\right)^3 * \left(\dfrac{1}{2}\right)^{(5 - 3)} = 10 * \dfrac{1}{8} * \dfrac{1}{4} = \dfrac{10}{32} \approx 31.25\%.$