Answer to Question #82257 in Algebra for Cartoon

Question

1. Suppose there are m Martians and n Earthlings at an intergalactic peace conference. To ensure the Martians stay peaceful at the conference, we must make sure that no two Martians sit together, such that between any two Martians there is always at least one Earthling.


(a) Suppose all m+n Martians and Earthlings are seated in a line. How many ways can the Earthlings and Martians be seated in a line?

(b) Suppose now that the m+n Martians and Earthlings are seated around a circular round-table. How many ways can the Earthlings and Martians be seated around the round-table?
And
2. How many positive integers does 6000 have?

Answer 1

1) Let the Martians and Earthlings are distinguishable among themselves.

a) Case one - n < (m - 1), then the possible arrangements of 0

(after the seating of all earthlings there will be <= m - 1 single places for Martians)

Case two - n> = (m - 1), then we will seed all over again all earthlings. The number of possible arrangements of earthlings n !,

places for n + 1 Martians (we can plant Martians from the edge).

Then the options for seating

(n + 1) * n * ... * (n - m + 2) = (n + 1)! / (n - m + 1)! (for the first Martian, all positions are free, for the second, 1 less)

Answer: n! * (n + 1)! / (n - m + 1)!

b) Case one - n < m, then possible arrangements of 0 (after seating all people, <= m - 1 single places for Martians will remain)

Case two - n> = (m - 1), we fix the 1st person at a specific place (we assume that the same arrangements when the rotation is the same),

the possible number of other people’s seating arrangements is (n - 1) !, places for Martians n.

Then the options to plant Martians - n * (n - 1) * ... * (n - m + 1) = n! / (n - m)!

(for the first Martian, all positions are free, for the second, 1 less)

Answer: (n - 1)! * n! / (n - m)!

2) 6000 = (2 ^ 4) * (5 ^ 3) * (3 ^ 1)

Then the number of divisors is (4 + 1) * (3 + 1) * (1 + 1) = 5 * 4 * 2 = 40