If a family will have $7$ children, how many combinations of boys and girls can there be if order doesn't matter? I feel like an idiot for asking this but I've been overthinking things. 
I know the answer is $8$, but without drawing a binomial tree diagram, I instinctively thought "$7$". 
What is the formula or mathematical thought process that people go through to get the $n+1$?
I totally understand that there would be $2^n$ combinations if the ordered mattered, but is there a way to logically reduce/divide that by the repeated combinations to arrive at $8$ as well? 
Thank god for anonymity and sorry if you're losing IQ points reading this. 
 A: The number of boys can be $0, 1, 2, 3, 4, 5, 6,$ or $7$. 
The corresponding number of girls is $7,6,5,4,3,2,1,$ or $0$. 
This means there are $\boxed{8\,}$ possibilities. 

The reason you intuitively think of $7$ is perhaps forgetting to account for one of the edge cases of $0$ boys or $0$ girls. 

Generalization from comment:

Suppose you are choosing $7$ fruits from a bowl. Each fruit can be an apple, pear, or banana. How many possible combinations of fruits are there, if order does not matter?

In this problem, we can use a technique called stars and bars. Imagine $7$ identical balls and $2$ identical dividers. Placing the $9$ objects in a line can be corresponded with a combination of apples, pears, and bananas in the original problem.
For each placement of dividers, we can say that the number of apples is the number of balls before the first divider. The number of pears is the number of balls between the first divider and the second divider. And finally, the number of bananas is the number of balls after the second divider.
The number of combinations of apples/pears/bananas exactly corresponds to the number of ways to arrange $2$ identical dividers and $7$ identical balls in a line. This is equal to the number of ways to choose the two places in which there are dividers. There are nine possibilities, of which we must choose two. Mathematicians call this "nine choose two" and it is written and computed as a binomial coefficient:
$$\binom{9}{2} = \frac{9 \cdot 8}{2} = \boxed{36\,}$$
A: If order doesn't matter, then each distinct combination is determined by the number of boys. And there can be anywhere from $0$ to $7$ boys, inclusive, so there are $8$ possible combinations.
