The probability of choosing the third card Five cards are numbered as the following
$$5 , 5 , 5 , 4 ,4$$
Three cards are chosen randomly , whats the probability that the third card is numbered 4 ? Without replacement
My turn
Let the first two cards are numbered 5 and the third is4
So we get
$$3C1 \times 2C1 \times 2C1$$
Let the first is 5 and the second and the third are 4
So 
$$3C1 \times 2C1 \times 1$$
 The same thing happens if the first and the third are 4 and the second is 5 so
The probability is $$\frac{ the sum }{5P3}$$
Is the solution correct ?
(Already answered in comments - there should be an Answer explaining carefully why this is trivial...)

The answer is obviously $2/5$; the first two cards chosen are irrelevant.

To be a little more formal: Let's give names to those five cards; say they're $C_1,\dots,C_5$ (so it happens that $C_1$, $C_2$ and $C_3$ are fives, while $C_4$ and $C_5$ are fours).

Say $X$ is the third card chosen. It's clear by symmetry that $$P(X=C_j)=P(X=C_k)$$for every $j$ and $k$; there's obviously no reason why the third card is more likely to be $C_2$ than $C_1$. Hence we have $$P(X=C_j)=1/5$$for every $j$, so $$P(X\text{ is a four})=P(X=C_4)+P(X=C_5)=2/5.$$
I have never liked calculating probability in terms of "binomial coefficients.  This is how I would do this problem:

There are four ways the first two cards can be drawn: 55, 54, 45, and 44.  

The probability the first card is a 5 is 3/5.  If the first card is a 5 then there are 4 cards left, 2 5s and 2 4s.  The probability of "55" is (3/5)(2/4)= 3/10.  In this case there are 3 cards left, one 5 and 2 4s, so the probability the third card drawn is a 4 is 2/3.  The probability of 554 is (3/10)(2/3)= 1/5.

The probability the first card is a 5 is 3/5.  If the first card is a 5 then there are 4 cards left, 2 5s and 2 4s. The probability of "54" is (3/5)(2/4)= 3/10. In this case there are 3 cards left, two 5s and one 4, so the probability the third card drawn is a 4 is 1/3.  The probability of "544" is (3/10)(1/3)= 1/10.

The probability the first card drawn is a 4 is 2/5.  If the first card is a 4 then there are 4 cards left, three 5s and one 4. The probability of "45" is (2/5)(3/4)= 3/5.  I this case there are 3 cards left, two 5s and one 4, so the probability the third card is a 4 is 1/3.  The probability of "454" is (3/5)(1/3)= 1/5.

Since there are only two 4s if the first two are "44" the third cannot be a 4.  The probability of "444" is 0.

Since these are independent events, the probability the third card is a 4 is 1/5+ 1/10+ 1/5= 2/10+ 1/10+ 2/10= 5/10= 1/2.

That is NOT the same as your $\frac{3C2+ 2C1+ 2C1}{5P3}= \frac{3+ 2+ 2}{60}= \frac{7}{60}$.