Answer:
5-card hands with at least one club: [tex] {52 \choose 5}-{39 ]choose 5}[/tex]
5-card hands with at least two cards of the same rank: [tex]{52 \choose 5}-{13 \choose 5}4^5[/tex]
Step-by-step explanation:
To determine how many 5-card hands have at least one club, we can count how many do NOT have at least one club, and then subtract that from the total amount of 5-card hands that there are.
A 5-card hand that doesn't have at least one club, is one whose 5 cards are from spades,hearts or diamonds. Since a standard deck of cards has 13 clubs, 39 of the cards are spades, hearts of diamonds. Getting a 5-card hand out of those cards, is choosing 5 cards out of those 39 cards. So there are [tex] {39 \choose 5}[/tex] 5-card hands without any clubs.
The total amount of 5-card hands is [tex]{52 \choose 5}[/tex], since a 5-card hand is simply a group of 5 cards out of the full deck, which has 52 cards.
Therefore the number of 5-card hands that have at least one club is [tex] {52 \choose 5}-{39 \choose 5}[/tex].
To determine how many 5-card hands have at least two cards with the same rank we can follow the same approach. We determine how many 5-card hands have NO cards with the same rank, and the subtract that out of the total amount of 5-card hands.
A 5-card hand that doesn't have cards of the same rank, is a group of 5 cards all from different ranks. Such hand can be made then by choosing first which 5 different ranks are going to be present in the hand, out of the 13 available ranks. So there are [tex] {13 \choose 5}[/tex] possible combinations of ranks. Then, choosing which card from each of the chosen ranks is the one that is going to be in the hand, is choosing which of 4 cards from EACH rank is going to be in the hand. So for each rank there are 4 availble choices, and so there are [tex]4^5[/tex] possible ways to choose the specific cards from each rank that will be in the hand. So the amount of 5-card hands with all ranks different is [tex] {13 \choose 5}\cdot{4^5}[/tex]
Therefore the amount of 5-card hands with at least two cards with the same rank is [tex]{52 \choose 5}-{13 \choose 5}\cdot4^5[/tex]
