A student draws 5 cards from a standard deck of 52 cards A student draws 5 cards from a standard deck of 52 cards


*

*How many ways could the selection result in a hand with no clubs?

*How many ways could the selection result in a hand at least 1 club?
I'm not sure I'm using pigeon-hole principle for both.
Even using the pigeon-hole, I have no idea how to apply on that.
 A: The are $52\cdot\frac{3}{4}=39$ cards which are not clubs. Part a) is effectively asking, given these 39 cards how many ways are there of choosing 5 in other words what is 39 choose 5:
$$\binom{39}{5}=575757$$
For part b) we can do something similar, lets start with choosing 1 club. There are 13 to choose from, then there are still 39 remaining non clubs for the remaining 4 cards so for hands with one club we have
$$13\cdot\binom{39}{4}$$
For hands with two clubs we have at first 13 to choose from, then 12 so we have $13\times12$ possibilities for the clubs. But this is just how many ways can we choose 2 cards from 13, i.e. $\binom{13}{2}$. Then for the non-clubs there are $\binom{39}{3}$ ways to choose the remaining 3 cards. Therefore for a hand involving $i$ clubs there are
$$\binom{13}{i}\cdot\binom{39}{5-i},$$
which we can sum from $i=1$ to $5$ 
$$\sum_{i=1}^5 \binom{13}{i}\cdot\binom{39}{5-i}.$$
According to Wolfram|Alpha is equal to 2,023,203 possible hands involving at least one club. 
Also rather than calculating this directly as above we can see that since there are $\binom{39}{5}$ ways of choosing a hand not involving a club then there must be $\binom{52}{5}-\binom{39}{5}$ ways of choosing a hand that has at least one club. The two answers agree (as one would hope!).
A: By direct counting: there are 13 clubs cards in a deck of 52 cards, the student has to draw 5 non-clubs cards:
he can pick the first non-clubs card out of 52-13=39 non-clubs cards;
he can pick the second non-clubs card out of 38 cards and so on.
Thus he can pick 5 non-clubs cards in 
$$
39 \cdot 38 \cdot 37 \cdot 36 \cdot 35 = \frac{39!}{(39-5)!}
$$
Since in the request of drawing 5 cards it doesn't matter the order the student picks these 5 cards, this number has to be divided by the number of ways you could arrange 5 cards: 5!
Thus he can pick 5 non-clubs cards in:
$$
 \frac{39!}{(39-5)!5!}={39 \choose 5}
$$
To me your question number 2 looks exactly like the number 1.
Edit: now I see the question has been edited.
For point 2 you can reason in the same way, but considering that the hand has to contain at least a club; i.e one card has to be choosen between 13 and the others between 52-1=51 cards. 
