So, what do we do? We select the best possible combination of foods to satiate our taste buds. There’s so much deliciousness on the menu which leaves us confused. Ordering food at a restaurant is never easy. So, keep reading! Real-life examples of permutations 1. We have jotted down some interesting examples in this write-up to help you understand how these math concepts find their way into the real world. On the contrary, combination involves arranging or selecting objects/ data from a large set, and the arrangement or order of selection does not matter. An important point to remember here is that the order of arrangement of objects/ data matters in permutation. Permutation involves arranging a set of objects or data in sequential order and determining the number of ways it can be arranged. But what exactly are they? While both terms are used together, they are not the same. There are several real-life situations where we use the knowledge we have learned in school about permutation and combination. 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 $C$.Would you believe it if we said that while playing the piano or making a cup of coffee, you’re unknowingly applying mathematical concepts of permutation and combination? Most definitely not. 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.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.
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