In a litter of 5 puppies, what is the probability that 2 males and 3 females are not obtained, or that 2 females and 3 males are not obtained? I have this statement:

In a litter of $5$ puppies, what is the probability that $2$ males and $3$ females are not obtained, or that $2$ females and $3$ males are not obtained?

My attempt was:
Make a Pascal's triangle to get the odds.
Let $M = Males, F = Females$
$1F^5M^0-5F^4M^1-10F^3M^2-10F^2M^3-5FM^4-1F^0M^5$
The probability of that $2$ males and $3$ females are not obtained is:
$1F^5M^0-5F^4M^1-\underbrace{10F^3M^2}_\text{Exclude this probability}-10F^2M^3-5FM^4-1F^0M^5$
$32/32 - 10/32 = 22/32$
Same procedure with the probability of that $2$ females and $3$ males are not obtained, that is $22/32$
Now i need the intersection that is $1F^5M^0 + 5F^4M^1 +5FM^4+1F^0M^5=12/32$ cases.
So, the probability is $22/32 + 22/32 - 12/32 = 32/32$. But this is wrong.
What is wrong with my development? I want to know where is my error and how to get an answer according my development. Thanks in advance.
 A: If you want to look strictly at Pascal's triangle, notice that the row adds up to $2^5=32$.  The numbers that you want to exclude are $10$ and $10$.  Therefore, you have
$$1-\frac{10+10}{32}=1-\frac{20}{32}=\frac{12}{32}=\frac38.$$
You have already counted females as non-males, and therefore, do not need to consider one case for males as the counter, then another case for females as the counter.  That is, you only need to consider that in $$\binom{5}{r}=\frac{5!}{r!\ (5-r)!}$$
for every $r$ males, you have $5-r$ females.
A: Let $F$ be the number of female puppies.  When you calculated $\Pr(F \neq 3)$, you found 
$$\Pr(F \neq 3) = \Pr(F = 0) + \Pr(F = 1) + \Pr(F = 2) + \Pr(F = 4) + \Pr(F = 5)$$
When you calculated $\Pr(F \neq 2)$, you found
$$\Pr(F \neq 2) = \Pr(F = 0) + \Pr(F = 1) + \Pr(F = 3) + \Pr(F = 4) + \Pr(F = 5)$$
When you calculated $\Pr(F \neq 2 \cap F \neq 3)$, you calculated
$$\Pr(F \neq 2 \cap F \neq 3) = \Pr(F = 0) + \Pr(F = 1) + \Pr(F = 4) + \Pr(F = 5)$$
Hence, 
\begin{align*}
\Pr(F \neq 2) & + \Pr(F \neq 3) - \Pr(F \neq 2 \cap F \neq 3)\\ & \quad = \Pr(F = 0) + \Pr(F = 1) + \Pr(F = 2) + \Pr(F = 3) + \Pr(F = 4) + \Pr(F = 5)\\
& \qquad = 1
\end{align*}
What you want to calculate is simply $\Pr(F \neq 2 \cap F \neq 3)$.
