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Zorluk: ZorPermutations, Combinations, and Fundamental Counting Principle

A logistics coordinator must assign 66 distinct delivery routes, labeled R1,R2,R3,R4,R5,R_1, R_2, R_3, R_4, R_5, and R6R_6, to 66 consecutive time slots, with exactly one route scheduled per slot. The schedule must satisfy two conditions:
1. Route R1R_1 must be scheduled in an earlier time slot than Route R2R_2.
2. Route R1R_1 and Route R2R_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.

  1. 240240, representing the total number of valid schedules satisfying both conditionsCevap
  2. 240240, representing the total number of schedules in which R1R_1 and R2R_2 are placed in consecutive time slots regardless of orderCevap
  3. 13\frac{1}{3}, representing the fraction of all possible unrestricted schedules that satisfy both conditionsCevap
  4. D
    360360, representing the total number of valid schedules satisfying both conditions
  5. E
    12\frac{1}{2}, representing the fraction of all possible unrestricted schedules that satisfy both conditions

Cevap

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!=7206! = 720. By symmetry, R1R_1 comes before R2R_2 in exactly half of these, or 360 permutations. Within these 360 permutations, those where R1R_1 and R2R_2 occupy consecutive slots treat (R1,R2)(R_1, R_2) as a single unit in fixed order, yielding 5!=1205! = 120 permutations. Thus, the number of valid schedules is 360−120=240360 - 120 = 240. Separately, the total number of schedules with R1R_1 and R2R_2 adjacent in any order is 2!×5!=2402! \times 5! = 240. Finally, the ratio of valid schedules to total schedules is 240/720=1/3240 / 720 = 1/3. Therefore, the options stating 240 total valid schedules, 240 adjacent schedules, and a 1/3 ratio are all correct.

Adım Adım Çözüm

1
Calculate the total number of unrestricted arrangements of the 6 routes.
Total unrestricted arrangements = 6!=7206! = 720.
6 distinct routes placed into 6 distinct slots can be ordered in 6!6! ways.
2
Apply the symmetry property to determine the number of schedules where R1R_1 comes before R2R_2.
Schedules with R1R_1 before R2=7202=360R_2 = \frac{720}{2} = 360.
In any permutation of distinct elements, R1R_1 is equally likely to appear before or after R2R_2.
3
Calculate the number of forbidden schedules where R1R_1 is immediately before R2R_2 (consecutive).
Forbidden schedules = 5!=1205! = 120.
Treat the ordered block (R1,R2)(R_1, R_2) as a single entity. Arranging this block alongside the remaining 4 routes gives 5!5! permutations.
4
Subtract forbidden schedules from the ordered schedules to find valid schedules.
Valid schedules = 360−120=240360 - 120 = 240.
Subtracting the consecutive cases from all cases where R1R_1 precedes R2R_2 satisfies both non-consecutive and ordering rules.
5
Determine the count of schedules where R1R_1 and R2R_2 are consecutive in any order, and calculate the overall valid ratio.
Adjacent schedules = 2!×5!=2402! \times 5! = 240; Valid ratio = 240720=13\frac{240}{720} = \frac{1}{3}.
Adjacent slots allow 2 internal orders for the block, yielding 240240. Comparing 240240 valid schedules to 720720 total gives a ratio of 13\frac{1}{3}.

Anahtar Kavram

Permutations with Adjacency Restrictions and Relative Order Symmetry
Tahmini Süre:2m 0s
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Konular

Permutations, Combinations, and Fundamental Counting Principle

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