Question

example 2
count cases
the card game euchre uses only the 9s, 10s, jacks, queens, kings, and aces. five - card hands are dealt to the players. how many euchre hands contain:
a) at least three queens?
b) at least two black cards?

Explanation:

Step1: Calculate total number of cards

There are 6 ranks (9s, 10s, jacks, queens, kings, aces) and 4 suits, so there are $6\times4 = 24$ cards in total. The combination formula is $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n$ is the total number of items and $r$ is the number of items to be chosen.

Step2: Calculate hands with at least three queens (part a)

Case 1: 3 queens

There are 4 queens. The number of ways to choose 3 queens is $C(4,3)=\frac{4!}{3!(4 - 3)!}=4$. The number of ways to choose the remaining $5-3 = 2$ cards from the remaining $24 - 4=20$ cards is $C(20,2)=\frac{20!}{2!(20 - 2)!}=\frac{20\times19}{2\times 1}=190$. So, the number of hands with 3 queens is $4\times190 = 760$.

Case 2: 4 queens

The number of ways to choose 4 queens is $C(4,4)=\frac{4!}{4!(4 - 4)!}=1$. The number of ways to choose the remaining $5 - 4=1$ card from the remaining 20 cards is $C(20,1)=\frac{20!}{1!(20 - 1)!}=20$. So, the number of hands with 4 queens is $1\times20=20$.
The number of hands with at least three queens is $760 + 20=780$.

Step3: Calculate hands with at least two black - cards (part b)

There are 12 black cards and $24-12 = 12$ red cards.

Case 1: 2 black cards and 3 red cards

The number of ways to choose 2 black cards from 12 is $C(12,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66$. The number of ways to choose 3 red cards from 12 is $C(12,3)=\frac{12!}{3!(12 - 3)!}=\frac{12\times11\times10}{3\times2\times1}=220$. The number of hands with 2 black and 3 red cards is $66\times220 = 14520$.

Case 2: 3 black cards and 2 red cards

The number of ways to choose 3 black cards from 12 is $C(12,3)=220$. The number of ways to choose 2 red cards from 12 is $C(12,2)=66$. The number of hands with 3 black and 2 red cards is $220\times66 = 14520$.

Case 3: 4 black cards and 1 red card

The number of ways to choose 4 black cards from 12 is $C(12,4)=\frac{12!}{4!(12 - 4)!}=495$. The number of ways to choose 1 red card from 12 is $C(12,1)=12$. The number of hands with 4 black and 1 red cards is $495\times12 = 5940$.

Case 4: 5 black cards

The number of ways to choose 5 black cards from 12 is $C(12,5)=\frac{12!}{5!(12 - 5)!}=792$.
The number of hands with at least two black cards is $14520+14520 + 5940+792=35772$.

Answer:

a) 780
b) 35772