Permutations in a Circle

<p><a href="https://www.prepswift.com/quizzes/quiz/prepswift-permutations-in-a-circle" target="_blank">Permutations in a Circle Exercise</a></p><p><strong><span style="color:#8e44ad;">Permutations in a Circle</span></strong> calculations are a bit counter-intuitive. Imagine we have three people: Abe, Bob, and Cob. If they were sitting in a row of three seats, six distinct arrangements are possible:</p> <p>$$ABC_1...ACB_2...BAC_3...BCA_4...CAB_5...CBA_6$$</p> <p style="text-align: center;"><em>And we can of course come to this number quite easily by simply doing the following:</em></p> <p>$$3 \times 2 \times 1 =6$$</p> <p><strong><span style="color:#e74c3c;">But What If...</span></strong></p> <p>...these three friends were sitting at a circular table with three seats around it? How many distinct arrangements would be possible? Can we simply do $3 \times 2 \times 1 =6$? The answer is no because, at a circular table, you get repeats. See below:</p> <center><svg height="168.4812469482422" style=" width:661.6000366210938px; height:168.4812469482422px; background: #FFF; fill: none; " width="661.6000366210938" x="0" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" y="0"> <svg class="role-diagram-draw-area" xmlns="http://www.w3.org/2000/svg"><g class="shapes-region" style="stroke: black; fill: none;"><g class="composite-shape"><path class="real" d=" M68,95.6 C68,62.68 94.68,36 127.6,36 C160.52,36 187.2,62.68 187.2,95.6 C187.2,128.52 160.52,155.2 127.6,155.2 C94.68,155.2 68,128.52 68,95.6 Z" style="stroke-width: 1; stroke: rgb(0, 0, 0); fill: none; fill-opacity: 1;"></path></g><g class="composite-shape"><path class="real" d=" M272,95.6 C272,62.68 298.68,36 331.6,36 C364.52,36 391.2,62.68 391.2,95.6 C391.2,128.52 364.52,155.2 331.6,155.2 C298.68,155.2 272,128.52 272,95.6 Z" style="stroke-width: 1; stroke: rgb(0, 0, 0); fill: none; fill-opacity: 1;"></path></g><g class="composite-shape"><path 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519L624 519C629 576 637 619 651 665C575 694 518 706 456 706C276 706 45 578 45 284Z" fill="rgb(74,144,226)" fill-opacity="1" stroke="rgb(74,144,226)" stroke-opacity="1" stroke-width="8" transform="matrix(0.024480000000000002,0,0,-0.024480000000000002,0,0)"></path></g></g></g></g><g><g><g><g transform="matrix(1,0,0,1,600.7999572753906,134.2800030517578)"><path d="M567 55C567 39 552 30 524 28L480 25L480 -3C582 0 582 0 602 0C622 0 622 0 724 -3L724 25L698 28C651 35 650 35 642 80L555 705L509 705L136 112C100 56 82 40 48 32L28 27L28 -3C120 0 120 0 140 0C159 0 161 0 247 -3L247 25L195 28C179 29 164 38 164 47C164 55 171 69 190 102L292 277L538 277L563 89L563 86C563 84 567 69 567 55ZM496 601L533 313L317 313Z" fill="rgb(74,144,226)" fill-opacity="1" stroke="rgb(74,144,226)" stroke-opacity="1" stroke-width="8" transform="matrix(0.024480000000000002,0,0,-0.024480000000000002,0,0)"></path></g></g></g></g><g><g><g><g transform="matrix(1,0,0,1,462.8000183105469,134.2800030517578)"><path d="M543 227C543 337 459 353 377 365C434 385 459 397 490 425C536 464 559 511 559 562C559 644 501 692 401 692C399 692 389 692 374 691L311 690C299 689 263 689 251 689C232 689 201 690 152 691L100 692L97 664L150 662C174 661 185 653 185 635C185 621 181 589 176 559L89 55C85 37 73 30 31 23L26 -3L63 -2C90 0 105 0 117 0C128 0 154 -1 180 -2L217 -3L237 -4C256 -5 269 -6 277 -6C392 -6 543 87 543 227ZM166 41L217 340L302 340C405 340 455 298 455 211C455 108 382 34 282 34C228 34 207 38 166 41ZM352 655C433 655 473 619 473 544C473 439 399 376 275 376L223 376L272 650C288 650 329 655 352 655Z" fill="rgb(74,144,226)" fill-opacity="1" stroke="rgb(74,144,226)" stroke-opacity="1" stroke-width="8" transform="matrix(0.024480000000000002,0,0,-0.024480000000000002,0,0)"></path></g></g></g></g><g><g><g><g transform="matrix(1,0,0,1,216.35623168945312,90.58999389648437)"><path d="M604 347L604 406L65 406L65 347ZM604 134L604 193L65 193L65 134Z" fill="rgb(99,181,0)" fill-opacity="1" stroke="rgb(99,181,0)" stroke-opacity="1" stroke-width="8" transform="matrix(0.03519,0,0,-0.03519,0,0)"></path></g></g></g></g><g><g><g><g transform="matrix(1,0,0,1,426.3562316894531,90.58999389648437)"><path d="M604 347L604 406L65 406L65 347ZM604 134L604 193L65 193L65 134Z" fill="rgb(99,181,0)" fill-opacity="1" stroke="rgb(99,181,0)" stroke-opacity="1" stroke-width="8" transform="matrix(0.03519,0,0,-0.03519,0,0)"></path></g></g></g></g></svg> </svg></center> <p>At a circular table these three cases are equal. I know what you&#39;re thinking. What do you mean equal? $A$ is at the top in the first case, $B$ is at the top in the second case, and $C$ is at the top in the third case.</p> <p>But that&#39;s not what we mean. We mean distinct arrangements in terms of who is sitting next to whom. Notice that, in <strong><u>all three cases</u></strong>, $A$ is to the right of $B$ and to the left of $C$. Notice that, in <strong><u>all three cases</u></strong>, $B$ is to the left of $A$ and to the right of $C$.</p> <p><strong><span style="color:#27ae60;">So How Do We Get Rid of the Repeats?</span></strong></p> <p>We use this formula below. Assuming that there are $n$ people or items around a circle arrangement, we do...</p> <p><span style="font-size:22px;">$$(n-1)!$$</span></p>