Rectangular Table, Six People

[800_or_bust]

Six people - A, B, C, D, E, and F - are being seated at a rectangular table with six seats. All of the seats are placed on the longest length of the table - three on one side, three on the other. No seats are placed on the short length of the rectangular table. If seats are selected at random, what is the probability that D is seated directly adjacent to F?

[DavidG@VeritasPrep]

Six people - A, B, C, D, E, and F - are being seated at a rectangular table with six seats. All of the seats are placed on the longest length of the table - three on one side, three on the other. No seats are placed on the short length of the rectangular table. If seats are selected at random, what is the probability that D is seated directly adjacent to F?

Nice question, complete with a well-crafted trap answer. (1/3 would have been another good trap.)

[GMATGuruNY]

Six people - A, B, C, D, E, and F - are being seated at a rectangular table with six seats. All of the seats are placed on the longest length of the table - three on one side, three on the other. No seats are placed on the short length of the rectangular table. If seats are selected at random, what is the probability that D is seated directly adjacent to F?

For D and F to be in adjacent seats, one of them must be in a CENTER SEAT.

Case 1: D in a center seat, with F in an adjacent seat
P(D is in a center seat) = 2/6. (Of the 6 seats, 2 are in the center.)
P(F is in an adjacent seat) = 2/5. (Of the 5 remaining seats, 2 are adjacent to D's center seat.)
To combine these probabilities, we multiply:
2/6 * 2/5 = 2/15.

Case 2: F in a center seat, with D in an adjacent seat
The resulting probability will be the same as in Case 1:
2/15.

Since a favorable outcome will be yield by Case 1 OR Case 2, we ADD the probabilities:
2/15 + 2/15 = [spoiler]4/15[/spoiler].

[DavidG@VeritasPrep]

Six people - A, B, C, D, E, and F - are being seated at a rectangular table with six seats. All of the seats are placed on the longest length of the table - three on one side, three on the other. No seats are placed on the short length of the rectangular table. If seats are selected at random, what is the probability that D is seated directly adjacent to F?

Here's how I thought about it:

The total number of possible arrangements is straightforward enough. 6 people and 6 seats, so 6!.

The number of desired outcomes is a little trickier. Imagine that D and F and sitting next to each other in the center and left seats on one side of the table. There'd be 4 more people to sit, and so 4! ways to place them. But D and F could also be in the center and right seats on that side of the table. So D and F can either sit center-left or center-right and there are two sides of the table, so there are 4 different places where D and F can sit. Last, we don't know if they're sitting D-F or F-D, so there are another 2! ways to sit. Desired: 4!*4*2!

Desired/Tota = (4!*4*2!)/6! = 4*2!/6*5 = [spoiler]4/15[/spoiler]

[DavidG@VeritasPrep]

(And note that 1/3 would have been the answer had there simply been 6 adjacent seats in a row - creating two rows of three is a nice wrinkle.)

[800_or_bust]

Six people - A, B, C, D, E, and F - are being seated at a rectangular table with six seats. All of the seats are placed on the longest length of the table - three on one side, three on the other. No seats are placed on the short length of the rectangular table. If seats are selected at random, what is the probability that D is seated directly adjacent to F?

Here's how I thought about it:

The total number of possible arrangements is straightforward enough. 6 people and 6 seats, so 6!.

The number of desired outcomes is a little trickier. Imagine that D and F and sitting next to each other in the center and left seats on one side of the table. There'd be 4 more people to sit, and so 4! ways to place them. But D and F could also be in the center and right seats on that side of the table. So D and F can either sit center-left or center-right and there are two sides of the table, so there are 4 different places where D and F can sit. Last, we don't know if they're sitting D-F or F-D, so there are another 2! ways to sit. Desired: 4!*4*2!

Desired/Tota = (4!*4*2!)/6! = 4*2!/6*5 = [spoiler]4/15[/spoiler]

Interesting solutions. I visualized this more as a strict probability question. There are six seats - four end seats and two center seats. Hence, there are two cases for which the probability would vary - D may be seated at an end seat, in which case F can must be seated on the center seat of the same side of the table; or D may be seated in a center seat, in which case F must be seated at one of the seats directly to the right or left of D.

Since there are four end seats out of six total, the probability that D is seated at an end seat is 2/3. And the probability that F is seated next to D, given that D is seated at an end seat is 1/5. Likewise, the probability that D is seated at a center seat is 1/3. And the probability that F is seated next to D, given that D is seated at a center seat, is 2/5.

Thus, the total probability is the sum of these individual probabilities: (2/3)(1/5) + (1/3)(2/5) = 4/15.

[800_or_bust]

Six people - A, B, C, D, E, and F - are being seated at a rectangular table with six seats. All of the seats are placed on the longest length of the table - three on one side, three on the other. No seats are placed on the short length of the rectangular table. If seats are selected at random, what is the probability that D is seated directly adjacent to F?

Here's how I thought about it:

The total number of possible arrangements is straightforward enough. 6 people and 6 seats, so 6!.

The number of desired outcomes is a little trickier. Imagine that D and F and sitting next to each other in the center and left seats on one side of the table. There'd be 4 more people to sit, and so 4! ways to place them. But D and F could also be in the center and right seats on that side of the table. So D and F can either sit center-left or center-right and there are two sides of the table, so there are 4 different places where D and F can sit. Last, we don't know if they're sitting D-F or F-D, so there are another 2! ways to sit. Desired: 4!*4*2!

Desired/Tota = (4!*4*2!)/6! = 4*2!/6*5 = [spoiler]4/15[/spoiler]

Interesting solutions. I visualized this more as a strict probability question. There are six seats - four end seats and two center seats. Hence, there are two cases for which the probability would vary - D may be seated at an end seat, in which case F can must be seated on the center seat of the same side of the table; or D may be seated in a center seat, in which case F must be seated at one of the seats directly to the right or left of D.

Since there are four end seats out of six total, the probability that D is seated at an end seat is 2/3. And the probability that F is seated next to D, given that D is seated at an end seat is 1/5. Likewise, the probability that D is seated at a center seat is 1/3. And the probability that F is seated next to D, given that D is seated at a center seat, is 2/5.

Thus, the total probability is the sum of these individual probabilities: (2/3)(1/5) + (1/3)(2/5) = 4/15.

The probabilistic approach also works in the case of a single, straight row, except in that case, the probability of D being seated in an end seat is only 1/3, whereas the probability that D is seated in a center seat is 2/3. Hence, in that case, the probability would be (1/3)(1/5) + (2/3)(2/5) = 1/3.

[Matt@VeritasPrep]

Seems like casework all the way:

If D is seated in the middle (p = 1/3), there's a (2/5) chance F is on either side of D on that side of the table.

If D is seated on an edge (p = 2/3), there's a (1/5) chance F is seated on the middle of the same side.

2 * (2/3) * (1/5) => 4/15