Counting: Permutations Question 5, Part A Solution
Things to look for in the problem: Shinobi was so delighted the mission was a success that he hosted a small gathering with his friends. When his eight friends arrived at his house, they decided they would spend the night playing draw poker. Shinobi’s square dinner table was piled too high with snacks so they decided to play poker on his round table . Upon the arrival of the two last guests, Shinobi received many questions about his independent mission and the bounty he recovered. He instinctively became defensive towards the two guests, but had no evidence against them and couldn’t throw them out of his house. Instead, he decided that he would sit next to each of the two guests as they played poker to watch over them like a hawk. *(Don’t mind the poker part yet, we will come back to that later.)
Because we are using a round table, we would use the formula for circular permutation which is represented by ( n -1)! . Where “ n ” is the number of objects to be arranged. We subtract 1 from the number of objects to be arranged (n) so it can act as a placeholder. Normally a circle does not have a start or stop point because it goes around forever. Therefore we need a placeholder to figure out where we should start counting and where we should end counting. We then arrange our other eight people around it in (n-1)! ways around our 1 placeholder. Our place holder.
In our question we have nine (9) people at the party, Shinobi and his eight guests. If we were to figure out how many ways they could be seated around the table our equation would look like this: There are 40320 ways to arrange nine people around a round table. However... (9-1)! = 8! = 40320 Arranging the nine people around the table subtracting 1 to be our placeholder. (n-1)! Our formula for circular permutations
However in our question we have this: If Shinobi insisted on sitting beside the two suspicious guests to keep an eye on them, how many ways can Shinobi and his eight guests be seated around the table? Shinobi insists on sitting next to the two suspicious guests. That means we MUST keep the three of them together. We would call this a case restriction where we can’t seat all nine people any old way. We have to find out how many ways we can arrange the people around the table while keeping Shinobi and the two suspicious people together at all times.
First we “tie” Shinobi and the two suspicious guests (3 people) together and they will count as one object right now. We then have the other six people to arrange around the table making six other objects to be arranged. So all together we have 7 objects to arrange around the table. We can now use the circular permutations formula. 1 2 3 4 5 6 7
Shinobi and the two suspicious guests. To be more visually accurate...
(n-1)! = (7-1)! = 6! = 720 7 being the number of objects to be arranged and we subtract 1 to be our placeholder. There are 720 ways to arrange 7 objects around a round table.
Our second step is to “untie” our 3 people. They themselves can be arranged 3! ways. 3! = 6 There are 6 ways to arrange 3 people.
Lastly, we multiply our two sets of arrangements by each other to get our total number of ways to arrange the people around the table. (7-1)! * 3! =720 * 6 = 4320 (7-1)! The number of ways to arrange 7 objects around the round table. And 3! The number that the three people who must stay together can be arranged.
There are 4320 ways for the nine people to be arranged around the table when Shinobi insists on sitting beside the two suspicious guests.