Difference between revisions of "User:Choi88" - Rhea

Latest revision as of 11:11, 1 October 2009

Guten Tag!! Ich konnte letztes Jahr nur englisch und franzozisch sprechen. Aber Ich habe im Sommer deutsch gelernt. Ich heiss "Joe" oder "Jungu" (auf Coreen). Ich bin Student an den Uni. Danke. 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)

Week 3:

Homework #46 in Chapter 5.5. In how many ways can a dozen books be placed on four distinguishable shelves: if no two books are the same, and the positions of the books on the shelves matter?

shelves: s1, s2, s3, and s4 books: b1, b2, ... , b11, b12

This question is similar to placing a number of distinguishable objects into distinguishable boxes (the same approach). Instead of considering shelves as s1, s2, s3, and s4, maybe we can think of them as bars with distinguishable labels.

for example:
 
 bar1, b1, bar2, b2 b3 b4, bar3, b5 b6 b7 b8, bar4, b9 b10 b11 b12

This is as trivial as lining 16 objects and the order does matter.
I find the answer to it as 16!