## electronicz Group Title At the head table at a banquet are seated two senators, two governors, and three mayors. Find the number of ways in which these seven people can be seated under the conditions described: A mayor is at each end and the senators are in consecutive seats one year ago one year ago

1. electronicz Group Title

hi

2. satellite73 Group Title

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3. satellite73 Group Title

lots of applications of the counting principle two choices for mayor

4. electronicz Group Title

ok so how will u do the senators?

5. satellite73 Group Title

oh i read it wrong, sorry

6. satellite73 Group Title

lets back up a bit

7. satellite73 Group Title

label the seats 1 to 7 then senators can be in seats 2, 3 3, 4 4, 5 5, 6 and they can switch chairs between them, so there are 8 possibilities for the senators

8. satellite73 Group Title

seat number 1 and 7 must contain mayors, so there are $$3\times 2=6$$ choices for seat 1 and 7

9. satellite73 Group Title

then you have 3 chairs left to fill , and 3 people left to sit, so they are $$3\times 2=6$$ choices for the remaining 3 chairs

10. satellite73 Group Title

multiply all these choices together, and unless i screwed up you get the right answer

11. electronicz Group Title

12. electronicz Group Title

@Mertsj ?

13. Mertsj Group Title

I do not see how the answer can only be 12.

14. Mertsj Group Title

Are you sure it is 12?

15. electronicz Group Title

well i guess order doesn't matter i nthnis case

16. electronicz Group Title

yea its 12... hey can u just help me on this question? how many committees of four can be chosen from twelve students? how many of these will include a given student? how many will exclude a given student?

17. Mertsj Group Title

It's a comination problem so 12 choose 4

18. Mertsj Group Title

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19. electronicz Group Title

yes what about the last two questions?

20. Mertsj Group Title

If we want to include a given student, we have already made one choice so we now must choose3 students from the remaining 11

21. electronicz Group Title

ok

22. Mertsj Group Title

So I think it would be 11 choose 3

23. Mertsj Group Title

I'm not really very good with probability.

24. satellite73 Group Title

you have to multiply all the choices together, as i wrote above

25. satellite73 Group Title

unless i screwed up, it should be $$8\times 6\times 6$$