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[[Category:MA375Spring2009Walther]]
 
[[Category:MA375Spring2009Walther]]
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Week 2
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I had some trouble with this question. I'd appreciate if anyone can add a comment on how I solved the problem.
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Prob 5.1: 40 How many ways can a photographer at a wedding arrange six people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if
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a) the bride must be in the picture?
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b) both the bride and groom must be in the picture?
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c) exactly one of the bride and the groom is in the picture?
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My solutions:
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a) If the bride is present in the photo, that leaves 9 people to choose from for the photo. There are 9 people to pick for the first position, 8 people for the second, 7 people for the third, 6 for the fourth, and 5 for the fifth. So it comes down to 9 * 8 * 7 * 6 * 5. However, since the bride can be present inbetween any of those 5 people, there are more possibilities. Say, a notion of a "gap" and a "space."
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  <space1> 1st <gap1> 2nd <gap2> 3rd <gap3> 4th <gap4> 5th <space2>.
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9 * 8 * 7 * 6 * 5 is the possibilities of lining up any 5 people. And the bride can be present in any of those gaps or spaces, which add up to 7. So the number of the total possibiilities boils down to 9 * 8 * 7 * 6 * 5 * 7
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b) This time we need to pick 4 people from a group of 8, excluding bride and groom. So it becomes 8 * 7 * 6 * 5.
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  <space1> 1st <gap1> 2nd <gap2> 3rd <gap3> 4th <space2>
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Bride and groom must be present in any of those spaces or gaps. there are 5 of them. So the number of possibilities becomes 5 * 4. And the total number of possibilities for b) is 8 * 7 * 6 * 5 * 5 * 4
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c) If bride is present, groom must be absent. If groom is present, bride must be absent.
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  (1) The number of possibilities that bride is present: 9 * 8 * 7 * 6 * 5 * 7
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  (2) The number of possibilities that groom is present: 9 * 8 * 7 * 6 * 5 * 7
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  (3) The number of possibilities that bride is present but not groom: (1) - the number of possibilities that both bride and groom are present
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  (4) The number of possibilities that groom is present but not bride: (2) - the number of possibilities that both bride and groom are present.
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  (5) The number of possibilities that both groom and bride are present: 8 * 7 * 6 * 5 * 4
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  (6) The number of possibilities that exactly one of the bride and the groom is in the photo: (3) + (4)
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  (3) = (1) - (5)
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  (4) = (2) - (5)
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  (3) + (4) = (1) + (2) - 2 * (5) = 2 * (9 * 8 * 7 * 6 * 5 * 7) - 2 * (8 * 7 * 6 * 5 * 4)

Revision as of 16:58, 29 January 2009

Week 2

I had some trouble with this question. I'd appreciate if anyone can add a comment on how I solved the problem.

Prob 5.1: 40 How many ways can a photographer at a wedding arrange six people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if a) the bride must be in the picture? b) both the bride and groom must be in the picture? c) exactly one of the bride and the groom is in the picture?

My solutions: a) If the bride is present in the photo, that leaves 9 people to choose from for the photo. There are 9 people to pick for the first position, 8 people for the second, 7 people for the third, 6 for the fourth, and 5 for the fifth. So it comes down to 9 * 8 * 7 * 6 * 5. However, since the bride can be present inbetween any of those 5 people, there are more possibilities. Say, a notion of a "gap" and a "space."

 <space1> 1st <gap1> 2nd <gap2> 3rd <gap3> 4th <gap4> 5th <space2>.

9 * 8 * 7 * 6 * 5 is the possibilities of lining up any 5 people. And the bride can be present in any of those gaps or spaces, which add up to 7. So the number of the total possibiilities boils down to 9 * 8 * 7 * 6 * 5 * 7

b) This time we need to pick 4 people from a group of 8, excluding bride and groom. So it becomes 8 * 7 * 6 * 5.

 <space1> 1st <gap1> 2nd <gap2> 3rd <gap3> 4th <space2>

Bride and groom must be present in any of those spaces or gaps. there are 5 of them. So the number of possibilities becomes 5 * 4. And the total number of possibilities for b) is 8 * 7 * 6 * 5 * 5 * 4

c) If bride is present, groom must be absent. If groom is present, bride must be absent.

 (1) The number of possibilities that bride is present: 9 * 8 * 7 * 6 * 5 * 7
 (2) The number of possibilities that groom is present: 9 * 8 * 7 * 6 * 5 * 7
 (3) The number of possibilities that bride is present but not groom: (1) - the number of possibilities that both bride and groom are present
 (4) The number of possibilities that groom is present but not bride: (2) - the number of possibilities that both bride and groom are present.
 (5) The number of possibilities that both groom and bride are present: 8 * 7 * 6 * 5 * 4
 (6) The number of possibilities that exactly one of the bride and the groom is in the photo: (3) + (4)
 (3) = (1) - (5)
 (4) = (2) - (5)
 (3) + (4) = (1) + (2) - 2 * (5) = 2 * (9 * 8 * 7 * 6 * 5 * 7) - 2 * (8 * 7 * 6 * 5 * 4)

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin