Question

You are dealt five cards from an ordinary deck of 52 playing cards. In how many ways can you get a full house and a five-card combination containing two jacks and three aces?

Answer 1

Hint: For the given problem, we have to use permutations and combinations concepts to solve this problem. First of all for housefull we have to find cases that contain 5 cards i.e. 3 cards from the same ordinal and 2 cards from different than the one we just choose from. Second we have to find cases for a five-card combination containing two jacks and three aces.

Complete step by step solution:
For the given Dec of cards, we are given to find how many ways you can get a full house and a five-card combination containing two jacks and three aces.
We can solve the problem using permutations and combinations.
To start, let's review what a standard deck of cards looks like: 13 ordinal cards (Ace, 2-10, Jack, Queen, King) - 1 of each ordinal in each of 4 suits (spades, clubs, hearts, diamonds), and so there are 52 cards.
The first thing we need is 3 cards of the same ordinal, so we can express that as taking 1 of the 13 ordinals and getting 3 of 4 of them:
\[\begin{align}
  & ^{13}{{C}_{1}}{{\times }^{4}}{{C}_{3}}=\dfrac{13!}{\left( 12! \right)\left( 1! \right)}\times \dfrac{4!}{\left( 3! \right)\left( 1! \right)} \\
 & \Rightarrow 13\times 4 \\
 & \Rightarrow 52 \\
\end{align}\]
The next thing we need is 2 cards from the same ordinal and this ordinal has to be different than the one we just choose from, so that looks like:
\[^{12}{{C}_{1}}{{\times }^{4}}{{C}_{2}}=\dfrac{12!}{\left( 11! \right)\left( 1! \right)}\dfrac{4!}{\left( 2! \right)\left( 2! \right)}\]
\[\begin{align}
  & \Rightarrow 12\times 4\times 3\times \dfrac{1}{2} \\
 & \Rightarrow 72 \\
\end{align}\]
And now we multiply them together:
\[52\times 72=3744\]
Therefore, there are 3744 ways to get full house.
For the number of hands we can draw getting specifically 2 Jacks and 3 Aces, we calculate that this way - we only need to concern ourselves with picking out the number of cards of the 4 available in each of the listed ordinals, and so we get:
\[\begin{align}
  & ^{4}{{C}_{3}}{{\times }^{4}}{{C}_{2}}=\dfrac{4!}{\left( 3! \right)\left( 1! \right)}\dfrac{4!}{\left( 2! \right)\left( 2! \right)} \\
 & =4\times 4\times 3\times \dfrac{1}{2} \\
 & =24 \\
\end{align}\]
So, therefore there are 24 ways of five-card combination containing two jacks and three aces.

Note: Students should be aware of permutations and combination concepts. They should be aware of calculation mistakes while doing this problem. They should be careful in every step while doing this problem because any mistake may change the value of the result.