A logistics coordinator must assign 6 distinct delivery routes, labeled R_1, R_2, R_3, R_4… — GRE General Test

A logistics coordinator must assign $6$ distinct delivery routes, labeled $R_1, R_2, R_3, R_4, R_5,$ and $R_6$ , to $6$ consecutive time slots, with exactly one route scheduled per slot. The schedule must satisfy two conditions:
1. Route $R_1$ must be scheduled in an earlier time slot than Route $R_2$ .
2. Route $R_1$ and Route $R_2$ cannot be scheduled in consecutive time slots.

Which of the following values correctly describe counts or proportions associated with this scheduling scenario? Select all that apply.

$240$ , representing the total number of valid schedules satisfying both conditionsAnswer
$240$ , representing the total number of schedules in which $R_1$ and $R_2$ are placed in consecutive time slots regardless of orderAnswer
$\frac{1}{3}$ , representing the fraction of all possible unrestricted schedules that satisfy both conditionsAnswer
D
$360$ , representing the total number of valid schedules satisfying both conditions
E
$\frac{1}{2}$ , representing the fraction of all possible unrestricted schedules that satisfy both conditions

Answer

The valid choices are the statement giving 240 as the total number of valid schedules, the statement giving 240 as the total number of schedules with adjacent routes, and the statement giving 1/3 as the fraction of valid schedules.

The total number of unrestricted permutations for 6 distinct routes is

6! = 720

. By symmetry,

R_1

comes before

R_2

in exactly half of these, or 360 permutations. Within these 360 permutations, those where

R_1

and

R_2

occupy consecutive slots treat

(R_1, R_2)

as a single unit in fixed order, yielding

5! = 120

permutations. Thus, the number of valid schedules is

360 - 120 = 240

. Separately, the total number of schedules with

R_1

and

R_2

adjacent in any order is

2! \times 5! = 240

. Finally, the ratio of valid schedules to total schedules is

240 / 720 = 1/3

. Therefore, the options stating 240 total valid schedules, 240 adjacent schedules, and a 1/3 ratio are all correct.

Step-by-Step Solution

Calculate the total number of unrestricted arrangements of the 6 routes.

Total unrestricted arrangements =

6! = 720

6 distinct routes placed into 6 distinct slots can be ordered in

6!

ways.

Apply the symmetry property to determine the number of schedules where

R_1

comes before

R_2

Schedules with

R_1

before

R_2 = \frac{720}{2} = 360

In any permutation of distinct elements,

R_1

is equally likely to appear before or after

R_2

Calculate the number of forbidden schedules where

R_1

is immediately before

R_2

(consecutive).

Forbidden schedules =

5! = 120

Treat the ordered block

(R_1, R_2)

as a single entity. Arranging this block alongside the remaining 4 routes gives

5!

permutations.

Subtract forbidden schedules from the ordered schedules to find valid schedules.

Valid schedules =

360 - 120 = 240

Subtracting the consecutive cases from all cases where

R_1

precedes

R_2

satisfies both non-consecutive and ordering rules.

Determine the count of schedules where

R_1

and

R_2

are consecutive in any order, and calculate the overall valid ratio.

Adjacent schedules =

2! \times 5! = 240

; Valid ratio =

\frac{240}{720} = \frac{1}{3}

Adjacent slots allow 2 internal orders for the block, yielding

240

. Comparing

240

valid schedules to

720

total gives a ratio of

\frac{1}{3}

Key Concept

Permutations with Adjacency Restrictions and Relative Order Symmetry

Estimated Time:2m 0s

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