![]() 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 $C$. 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. Learn about factorial, permutations, and combinations, and look at how to use these ideas to find probabilities. I would argue that the order of these two chosen values is not relevant and therefore calls for a $C$ to be calculated. About this unit How many outfits can you make from the shirts, pants, and socks in your closet Address this question and more as you explore methods for counting how many possible outcomes there are in various situations. 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 $C$). It contains a few word problems including one associated with the. 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. Join Subscribe 2M views 6 years ago New Precalculus Video Playlist This video tutorial focuses on permutations and combinations. ![]() 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. ![]() As per the fundamental principle of counting, there are the sum rules and the product rules to employ counting easily. Permutations are understood as arrangements and combinations are understood as selections. This leads immediately to the thought of $C(4,1)$ for five cards yiedling the given answer of $C(13,5)*(C(4,1))^5$. Permutation and combination are the methods employed in counting how many outcomes are possible in various situations. 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. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. As such, to choose $5$ distinctly valued cards need employ $C(13,5)$. What are permutations and combinations A permutation is an arrangement in a definite order of a number of objects taken, some or all at a time. 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: Note that each number in the triangle other than the 1's at the ends of each row is the sum of the two numbers to the right and left of it in the row above.The other two answers correctly mentioned the difference between order mattering and not mattering for permutations and combinations respectively. The triangular array of binomial coefficients is called Pascal's triangle after the seventeenth-century French mathematician Blaise Pascal.
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