Question

Probability and basic counting principles:
In a conference, Dr. Bond's lecture is related to Dr. Hernandez' lecture and should not precede it. If there are six more speakers, how many orderings of the eight speakers are possible? (work to follow...)

Answer 1

Good problem!

Answer 2

So how far did you get with this?

Answer 3

With 8 speakers, there are 8! orderings. I'm not sure the best way to calculate this, so I created a table, with H being Dr. Hernandez' lecture and numbers representing the number of choices per speaking slot.

6 5 4 3 2 1 H 1 = 6!
6 5 4 3 2 H 2 1 = 6! * 2
6 5 4 3 H 3 2 1 = 6! * 3
6 5 4 H 4 3 2 1 = 6! * 4
6 5 H 5 4 3 2 1 = 6! * 5
6 H 6 5 4 3 2 1 = 6! * 6
H 7 6 5 4 3 2 1 = 7!

Thus, I have 6*6!*6!*7!, or 8!/2, or 20,160 possible outcomes. a), is this correct and b) is there an easier way to solve it?

Answer 4

Are you familiar with the forumula for permutations without repetition?

Answer 5

nPr = n!/r! ?

Answer 6

Nearly.

nPr = n!/(n-r)!

Answer 7

Yes, that one. :)

Answer 8

Right. And are you familiar with combinations (without repetition)?

Answer 9

Yes, nCr = n!/r!(n-r)!

Answer 10

That's right. One second though. I think I made an assumption about the problem let me check something.

Answer 11

Probability and basic counting principles:
In a conference, Dr. Bond's lecture is related to Dr. Hernandez' lecture and should not precede it. If there are six more speakers, how many orderings of the eight speakers are possible? (work to follow...)

H.B.6.5.4.3.2.1 is what I get if I read it right

Answer 12

8!/2

Answer 13

blah. Checking something (again)

Answer 14

Here would be my approach:
Using your table, you see there are 7 places Dr H can fill. Each of these has a different number of places Dr B can fill (1, 2, 3, 4, etc.) depending on Dr H's position.

Regardless of Dr H and Dr B's positions, the remaining 6 can be placed 6! ways.

So, set it up for each row...
Row 1. There is 1 place for Dr B. so there are 1*6! ways to place Dr B and the other 6 speaker.

Row 2. There are 2 places for Dr B. so there are 2*6! ways to place Dr. B and the 6 others.

Continue with this and then add them up for the 7 ways to place Dr. H.

This give 1*6! + 2*6! + ... + 7*6! = 6!(1 + 2 + ... + 6 + 7)

Answer 15

1 + 2 + ... + 7 = 28, so you'd have 28*6! possibilities

Answer 16

there are 8! ways to arrange the speakers. Half will have B before H and half will have H before B

Answer 17

Hey polpak once u're done here can u look over this one problem.. I did most of it, I just need to know how to finish it..

Answer 18

amistrec: H doesn't have to go first.
mtbender: Which, I believe is the same as what I did.

Answer 19

Zarkon, it is so simple when you put it that way.

Answer 20

Thanks everyone.

Answer 21

I just placed H and B someplace where theyd be out of the way ... then counted the rest of them ...

Answer 22

Good observation Zarkon. By requiring B to go after H we remove half of our options.

Answer 23

itis possible a simply fail to understand the question :)

Answer 24

Logically, dividing by 2 makes sense. Is there a way to prove that other than as I and mtbender have done?

Answer 25

The problem is, I could never see the obvious, simple solution on tests...that's why I default to chipping away at it :)

Answer 26

Hey, I'm with you mtb. ;)

Answer 27

A basic proof would be if you extended your rows one more to allow H to be last, you can say that each row has a partner that's identical to it. Namely, the first and last rows are the same, 2nd and 7th are the same, etc.

You're simply looking at the line of speakers in reverse.

So the top 4 of these rows have B happening before H so you don't need to deal with them since you already counted it's pair...in reverse.