Combination vs permutation11/19/2023 ![]() ![]() The combination can also be random in nature. Values or objects in a combination do not require order or arrangement. One combination comprises one value plus another value (as a pair) with or without additional values (or as a multiple). In a combination, the importance is made on the choice of the objects or values themselves. “Combination” is defined as the selection of objects, symbols, or values from a wide variety like a large group or a certain set with underlying similarities. As mathematical concepts, they serve as precise terms and language to the situation they are describing or covering. Because they are related concepts, most of the time they are used with each other or switched or swapped with each other without realizing it. 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 $C$.Permutations and combinations are both related mathematical concepts. 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 $C(4,3)*C(4,2)$. 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. 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. ![]()
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