#COMBINATION VS PERMUTATION HOW TO#
#COMBINATION VS PERMUTATION FULL#

This is like saying 'we have r + (n−1) pool balls and want to choose r of them'. The same set of objects, but taken in a different order will give us different permutations.

A permutation pays attention to the order that we select our objects. So (being general here) there are r + (n−1) positions, and we want to choose r of them to have circles. What is the difference between a combination and permutation The key idea is that of order. Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (we need to move 4 times to go from the 1st to 5th container). So instead of worrying about different flavors, we have a simpler question: 'how many different ways can we arrange arrows and circles?' Let's use letters for the flavors: (one of banana, two of vanilla): This culminates in a solution of $C(13,2)P(2,1)C(4,3)C(4,2)$ which of course is the same value as the OP's just arrived at in a, in my opinion, clearer way as it stays a bit truer to the definitions that are the answer to the over-arching question of what distinguishes a $P$ from a $C$.Let us say there are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. Then finally we can zoom in individually on our pair and $3$-of-a-kind and choose our suits $3$ of them for our $3$-of-a-kind choice and $2$ for our pair yielding $C(4,3)*C(4,2)$.

#COMBINATION VS PERMUTATION FULL#

Because the choice of these two determines which value will require $3$ cards and which $2$ in our full house we are effectively distinguishing between a full house of $(A,A,K,K,K)$ and $(A,A,A,K,K)$ with this permutation of $P(2,1)$. the $2$ values just chosen in $C(13,2)$, to choose the grouping of $3$ and $1$ option for the grouping of $2$. We care not that we chose $(A,K)$ versus $(K,A)$ After this choice, we can ask now how many ways can I pick my grouping of $3$ and my grouping of $2$? Well one would have $2$ options, i.e. $C(13,2)$ is the clearer way because, like the $5$-of-a-kind example we are merely choosing cards. The note above's arguement would have you say that not only are you choosing two of the $13$ values for your full house but you are simultaneously choosing which will be your group of $3$ and which your group of $2$, and though this is technically correct I believe it skips a step and is the reason the OP was confused. I would argue that the order of these two chosen values is not relevant and therefore calls for a $C$ to be calculated. And I agree that it gives the correct value I just argue that it is arrived at via misleading reasoning based on the order mattering definitions differentiating $P$ and $C$). So exaclty like in the $5$-of-a-kind example we can choose $2$ of the $13$ possible values in a standard deck of $52$ cards to determine the number of ways in which choosing these two values is possible (NOTE that some may disagree that clearly there are $13$ options for the first value in a full house and $12$ for the other and thus $P(13,2)$ is correct. We know that a full house consists of $2$ values of card, $2$ of the one and $3$ of a second. I actually quite dislike the answer given in the OP, that of $P(13,2)C(4,2)C(4,3)$ as I am of the impression that it is hiding the real permutation in the $P(13,2)$ term.

This leads immediately to the thought of $C(4,1)$ for five cards yiedling the given answer of $C(13,5)*(C(4,1))^5$.

Now however, we have to realize that each card can come in four different suit, and thus for each of the five cards we chose with $C(13,5)$ we have $4$ options to choose from. As such, to choose $5$ distinctly valued cards need employ $C(13,5)$. a hand of $A,2,3,4,5$ all fo hearts is identical to $5,4,3,2,A$ all of hearts.

#COMBINATION VS PERMUTATION HOW TO#

How to find the total number of full houses in a poker handįor the first, as was previous mentioned, the order of the hand of five distinctly valued cards does not change the hand we have i.e.

How to find the total number of $5$-of-a-kinds in a poker hand.

I however, wanted to shed some light on the implied problems of the OP which were: The other two answers correctly mentioned the difference between order mattering and not mattering for permutations and combinations respectively.

Combination vs permutation