Lesson Persons sitting around a cicular table
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<H2>Persons sitting around a circular table</H2> <H3>Problem 1</H3>Snow White and the Seven dwarves sat around a circular table. How many seating arrangements were possible? <B>Solution</B> <pre> There are 7 dwarfs and 1 Snow White for a total of 8. If you are a novice in such problems, you may think that the answer is 8! It is not so ! The correct answer is 7! = 7*6*5*4*3*2*1 = 5040. Why ? - Because for circular arrangements "around a circular table" we consider two arrangements as identical (indistinguishable) if one arangements is the second turned as a whole thing around the circle ("around the table"). Therefore, each single arrangements of "n" things around the table generates n other dispositions that are considered as the same arrangement. Therefore, when consider circular arrangements, the number of all permutations n! must be divided by "n", which leaves only {{{n!/n}}} = (n-1)! different arrangements. </pre> <H3>Problem 2</H3>In how many ways can 20 board members sit in a circular board room if the chairman must sit between the secretary and the treasure? <B>Solution</B> <pre> In problems like this one, the two arrangements are considered as undistinguished (as one unique arrangement) if there is a rotation which maps one arrangement to the other. Based on this agreement, the number of different arrangements is 2*(20-3)! = 2*17! The factor "2" describes two different placements for the secretary and the treasure. 17! is the number of different arrangements for the rest 20-3 = 17 members. </pre> <H3>Problem 3</H3>In How many ways can 11 people be seated around a circular table if two of them insist th sit next to each other? <B>Solution</B> <pre> We consider this special pair as one (glued) object, and we, actually, have 10 objects then (instead of 11) to arrange around the circular table. We can arrange 10 objects around the circular table by 9! ways, but this special object can be in one of the two states (AB) or (BA). Therefore, the total number of all possible arrangements of this kind is 2*9! = 2 * (9*8*7*6*5*4*3*2*1) = 725760. <U>ANSWER</U> </pre> <H3>Problem 4</H3>In how ways can 5 boys and 3 girls be arranged in a circle if the girls must stand together? <B>Solution</B> <pre> Since the boys and the girls are arranged in a circle, we can think that the girls go first ("women first") and we arrange and count our circular permutations starting from the girls. We can start counting (arranging) from any of the 3 girls, arranging them in 3! = 6 ways. Then we arrange 5 boys by 5! = 120 ways (120 permutations for 5 objects). That's all. In all, we have 6*120 = 720 circular arrangements. Notice that we just accounted for the circular symmetry, when started our arrangements from the girls. So, the answer is 6*120 = 720 ways/(circular arrangements). </pre> <H3>Problem 5</H3>In how many ways can 4 gentlemen and 2 ladies be seated at a round table so that the ladies are not together? <B>Solution</B> To solve this problem, the best way / (the standard way) is to consider the whole set of permutations and the complementary set of permutations, and then to take the difference. <pre> 1. The whole set of permutations in this problem is the set of all permutations (seating arrangements) of 6 persons at a round table without considering gender. It is well known fact that the number of all such permutations (circular permutations) is (6-1)! = 5! = 120. 2. The complementary set of permutations is the set, where two ladies are sitting together. By considering this pair as one object, we have then the set of all circular permutations of 5 objects, which consists of (5-1)! = 4! = 24 permutations. We then must double this number 2*24 = 48 to distinct permutations of the type (Alice-Beatrice) and (Beatrice-Alice) inside these pairs. It gives the final answer 120 - 48 = 72. <U>Answer</U>. In how many ways can 4 gentlemen and 2 ladies be seated at a round table so that the ladies are not together? - in 72 wys. </pre> <H3>Problem 6</H3>In how many ways can 7 charms be placed in a bracelet which has no clasp? <B>Solution</B> <pre> From the dictionary <A HREF=https://dictionary.cambridge.org/us/dictionary/english/bracelet>https://dictionary.cambridge.org/us/dictionary/english/bracelet</A> https://dictionary.cambridge.org/us/dictionary/english/bracelet you can read that a bracelet is a piece of jewelry that is worn around the wrist or arm. In this definition, it is important for us now that a bracelet has a circular form like a closed line. So, the question can be EQUIVALENTLY reformulated in <U>THIS WAY</U> In how many ways can 7 charms be placed along a circumference of a circle ? In such problems the placements that obtained one from the other by rotation of a circle by some angle are considered as INDISTINGUISHABLE. Therefore, with each concrete placements, 6 others that obtained from the original placement by rotation, are considered as EQUIVALENT. Therefore, the number of all possible placements of 7 charms in a bracelet is {{{7!/7}}} = 6! = 1*2*3*4*5*6 = 720. <U>ANSWER</U> It is not 7!, as it would be in linear case ! </pre> My lessons on Permutations and Combinations in this site are - <A HREF =http://www.algebra.com/algebra/homework/Permutations/Introduction-to-Permutations.lesson>Introduction to Permutations</A> - <A HREF =http://www.algebra.com/algebra/homework/Permutations/PROOF-of-the-formula-on-the-number-of-permutations.lesson>PROOF of the formula on the number of Permutations</A> - <A HREF =http://www.algebra.com/algebra/homework/Permutations/Problems-on-Permutations.lesson>Simple and simplest problems on permutations</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Special-type-permutations-problems.lesson>Special type permutations problems</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/How-many-different-permutations-may-exist-ubder-given-restrictions.lesson>Problems on Permutations with restrictions</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Math-circle-level-problem-on-Permutations.lesson>Math circle level problem on Permutations</A> - <A HREF =http://www.algebra.com/algebra/homework/Permutations/Introduction-to-Combinations-.lesson>Introduction to Combinations</A> - <A HREF =http://www.algebra.com/algebra/homework/Permutations/PROOF-of-the-formula-on-the-number-of-combinations.lesson>PROOF of the formula on the number of Combinations</A> - <A HREF =http://www.algebra.com/algebra/homework/Permutations/Problems-on-Combinations.lesson>Problems on Combinations</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Combinations-problems-with-restrictions.lesson>Problems on Combinations with restrictions</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Math-circle-level-problem-on-Combinations.lesson>Math circle level problem on Combinations</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Arranging-elements-of-sets-containing-undistinguishable-elements.lesson>Arranging elements of sets containing indistinguishable elements</A> - Persons sitting around a circular table (this file) - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Combinatoric-problems-for-entities-other-than-permutations-and-combinations.lesson>Combinatoric problems for entities other than permutations and combinations</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Fundamental-counting-principle-problems.lesson>Fundamental counting principle problems</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Miscellaneous-problems-on-permutations-combinations-and-other-combinatoric-entities.lesson>Miscellaneous problems on permutations, combinations and other combinatoric entities</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Some-twisted-combinatorics-problem.lesson>Some twisted combinatorics problem</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Inclusion-Exclusion-principle.lesson>Inclusion-Exclusion principle problems</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/DERANGEMENT-problems.lesson>Derangement problems</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/In-how-many-ways-N-distinguishable-objects--can-be-distributed-among-n-different-boxes.lesson>In how many ways N distinguishable objects can be distributed among n different boxes ?</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Stars-and-bars-method-for-Combinatorics-problems-2.lesson>Stars and bars method for Combinatorics problems</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Men-and-women-standing-in-line-.lesson>Men and women standing in line</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/One-combinatorial-Geometry-problem-solved-using-the-Euler-formula.lesson>One combinatorial Geometry problem solved using the Euler formula</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/Nice-recreational-problems-on-permutations.lesson>Nice recreational problems on permutations</A> - <A HREF =https://www.algebra.com/algebra/homework/Permutations/OVERVIEW-the-lessons-on-Permutations-and-Combinations.lesson>OVERVIEW of lessons on Permutations and Combinations</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-II.
