Probability, Why is this wrong? So we have a deck of 32 cards , let's suppose we take 5 cards at the same time,
then we search for the probability of having at least one ♥ heart card
for this we can do 1- p(having no heart card) which translate to 
1 - (24 C 5)/(32 C 5)
which will give you approximately 0.77
so far it's all good,
the problem start here : 
we can (or can't?) also consider that "having at least one heart" as : "pick one heart then any other 4 cards"
which translate (or not?) to :
((8 C 1)*(31 C 1))/|Ω|
which give 1.25, which is obviously wrong
The question is, where is the mistake is my process?
 A: You overcount the hands with more than one heart in the second method.
If the hand contains, for example, both the ace and king of hearts and three non-hearts, such hands are counted twice, because you may have first selected the heart ace as the heart, and the king among the 4 out of remaining 31. Or the other way round. Should a hand contain more than two hearts the overcounting becomes more severe, but those hands are not very numerous.
A correct variant is the following.
Assume that the hand has $k$ hearts and hence $5-k$ non-hearts. The number of such hands is clearly
$$
{}_8C_k\cdot{}_{24}C_{5-k}=\binom 8 k\cdot\binom{24}{5-k}
$$
depending on how you want to write the binomial coefficients.
Thus the number of hands with at least one heart is
$$
\sum_{k=1}^5{}_8C_k\cdot{}_{24}C_{5-k}=158872.
$$
As a reality check we see that the total number of hands is $\binom{32}5=201376$, and the number of hands with no hearts $\binom{24}5=42504=2013776-158872.$
A: You neglected to mention that there were eight $\heartsuit$ among the 32 cards in the deck.
Now the probability of drawing at least one $\heartsuit$ can be calculated by complements $1-P(H=0)$, directly, or by using the Principle of Inclusion and Exclusion.
$$\begin{align}P(H{=}0) & = \dfrac{\dbinom{24}{5}}{\dbinom{32}{4}} \\[2ex] P(H{\geq} 1) & = \dfrac{\dbinom{8}{1}\dbinom{24}{4}+\dbinom{8}{2}\dbinom{24}{3}+\dbinom{8}{3}\dbinom{24}{2}+\dbinom{8}{4}\dbinom{24}{1}+\dbinom{8}{5}}{\dbinom{32}{4}}\\[2ex] P(H{\geq} 1) & = \dfrac{\dbinom{8}{1}\dbinom{31}{4}-\dbinom{8}{2}\dbinom{30}{3}+\dbinom{8}{3}\dbinom{29}{2}-\dbinom{8}{4}\dbinom{28}{1}+\dbinom{8}{5}}{\dbinom{32}{4}}\end{align}$$
You attempted the latter, but only used the first term, which over counts cases where more hearts are drawn.
