Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

lots of applications of the counting principle two choices for mayor

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

ok so how will u do the senators?
oh i read it wrong, sorry
lets back up a bit
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
seat number 1 and 7 must contain mayors, so there are \(3\times 2=6\) choices for seat 1 and 7
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
multiply all these choices together, and unless i screwed up you get the right answer
... the answer is 12
I do not see how the answer can only be 12.
Are you sure it is 12?
well i guess order doesn't matter i nthnis case
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?
It's a comination problem so 12 choose 4
yes what about the last two questions?
If we want to include a given student, we have already made one choice so we now must choose3 students from the remaining 11
So I think it would be 11 choose 3
I'm not really very good with probability.
you have to multiply all the choices together, as i wrote above
unless i screwed up, it should be \(8\times 6\times 6\)

Not the answer you are looking for?

Search for more explanations.

Ask your own question