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More examples on perms and combs

More examples on perms and combs. exercise. 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 groom are among those people, such that (a) the bride is in the picture (b) the bride and the groom are in the picture

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More examples on perms and combs

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  1. More examples on perms and combs

  2. exercise 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 groom are among those people, such that (a) the bride is in the picture (b) the bride and the groom are in the picture (c) either the bride is in the picture or the groom is in the picture, but not both

  3. exercise 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 groom are among those people, such that (a) the bride is in the picture • The bride can be in one of 6 positions • we then have 5 positions to permute from 9 people • P(n,r) where n=9 and r=5 • 9!/(9-5)! • 9.8.7.6.5 • therefore there are 6.9.8.7.6.5 ways to do this • 90720

  4. exercise 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 groom are among those people, such that (b) the bride and the groom are in the picture • The bride can be in one of 6 positions and the groom in one of 5 (or vice versa) • 30 alternatives in all • we then have 4 positions to permute from 8 people • P(n,r) where n=8 and r=4 • 8!/(8-4)! • 8.7.6.5 • therefore there are 30.8.7.6.5 ways in total • 50400

  5. exercise 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 groom are among those people, such that (c) either the bride is in the picture or the groom is in the picture, but not both • consider the bride without the groom • the bride can be in one of 6 positions • the remaining 5 positions can be taken by 8 people • i.e. 8 guests, not the groom! • P(n,r) where n=8 and r = 5 • 8!/(8-5)! • 8.7.6.5.4 • in total there are 6.8.7.6.5.4 ways to have the bride without the groom • 40320 ways • The same holds for the groom without the bride • therefore 40320 + 40320 = 80640 ways in total

  6. exercise In 3d year, single honours, students must take 9 modules with 9 exams (I’m ignoring the team project in this discussion). Each module has the following exam structure How many different ways can you satisfy the above requirements?

  7. fin

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