So, two places are occupied. In how many ways can 4 people occupy 10 chairs in a row if no two sit on adjacent chairs. Should I speak up for her? What wording of the question says I am looking for permutations and not combinations? The placement of this problem in the book doesn't make complete sense. Course Hero is not sponsored or endorsed by any college or university. In how many ways can these people be arranged for a photograph? The same reasoning is used to solve b) (permutations again), which comes to $1 \times 1 \times 8 \times 7 \times 6 \times 5 = 1,680$ combinations times $6 * 5$ ways to permute each combination, so the answer is $30 * 1680 = 50,400$. how to append public keys to remote host instead of copy it, Suggestions for braking with severe osteoarthritis in both hands. Counterpart to Confidante: Word for Someone Crying out for Help. A type of compartment that rises out of a desk, Telling my supervisor about my medical condition, In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, where the bride and the groom are Thanks for contributing an answer to Mathematics Stack Exchange!
In how many ways may 8 people form a circle for a folk dance? $8\cdot7\cdot6\cdot5\cdot4=6720$ ways to choose and arrange the other five people. The objective is to determine the number of ways can a photographer arrange six people in a row, including the bride and groom at a wedding. a row from a group of 10 people where the bride and groom are part of the group of 10 such that (a) at least one of them (the bride and the groom) must be in the pic? For example, to solve part a), I calculated all the (unique) combinations of 6 people from a group of 10 where 1 is always the bride. there is no restriction about their position so can be arranged 6! I am very confused by the wording of this question from Kenneth Rosen's Discrete Mathematics 6th Edition. The bride and the groom are the priority so they should be the most prominent. How many ways can a photographer at a wedding arrange 6 people in. Thank you. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. This problem comes after chapter 5.1, which is an introduction to the basic principles of counting: product rule, sum rule, and inclusion-exclusion principle. How easy is it to recognize that a creature is under the Dominate Monster spell? Bride and groom take two places each of 1 way.
$2$ ways to choose exactly one of the bride and groom. b) both the bride and the groom must be in the picture? © 2003-2020 Chegg Inc. All rights reserved. If we were considering permutations, would it not need to be $[6 * 40320] + [6 * 40320] = 483,840$ permutations? My wife's contributions are not acknowledged in our group's paper that has me as coauthor. c) exactly one of the bride and the groom is in the picture? Terraforming Mars using a combination of aerogel and GM microbes? MathJax reference. Therefore I would not have access to that knowledge directly. Since you are choosing from the bride or the groom, the answer is, One person from bride and groom can be selected by 2c1 way. It only takes a minute to sign up. However, this is just combinations, and not permutations to my understanding, because there are 6 ways to arrange either the bride or groom for each such combination. However, part c) is incredibly confusing because the answers I have found seem to just use combinations, and not permutations. However, this is just combinations, and not permutations to my understanding, because there are 6 ways to arrange either the bride or groom for each such combination. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Should I use constitute or constitutes here? Kenneth Rosen's Discrete Mathematics 6th Edition, Creating new Help Center documents for Review queues: Project overview. Different ways of arranging a group of 10 people. Thus, there are $80640=(2\cdot6)\cdot8\cdot7\cdot6\cdot5\cdot4$ permutations in this case.
To learn more, see our tips on writing great answers. The reasoning you used in parts (a) and (b) is correct. How do you win a simulated dogfight/Air-to-Air engagement? This comes to $1 \times 9 \times 8 \times 7 \times 6 \times 5 = 15,120$. In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if the bride must be in the picture? Discrete Mathematics and Its Applications | 6th Edition, Discrete Mathematics and Its Applications, 9780072880083, 9780073107790, 9780073312712. Now we need 5 more persons = 8c5 For some reason I thought the number of combinations was 40320 and not 6720!
Yet, the answers I have found simply add both combinations together: $40320 + 40320 = 80,640$. Why does the wording of how many ways can a photographer 6 people from a group of 10 ask for permutations and not combinations?
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