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Zorluk: ZorCounting with Restrictions and Repetitions

A conference schedule consists of 66 consecutive time slots. The organizers must schedule 33 identical workshops on Artificial Intelligence, 22 identical workshops on Cybersecurity, and 11 keynote address on Data Privacy. If the 22 Cybersecurity workshops cannot be scheduled in consecutive time slots, how many distinct presentation schedules are possible?

  1. A
    20
  2. 40Cevap
  3. C
    50
  4. D
    60
  5. E
    120

Cevap

40 distinct presentation schedules are possible.
The total number of unrestricted ways to arrange the 6 events (3 identical AI, 2 identical Cybersecurity, 1 Data Privacy) is calculated using multiset permutations as 6!3!×2!×1!=60\frac{6!}{3! \times 2! \times 1!} = 60. To find the number of ways where the two Cybersecurity workshops are NOT consecutive, we use complementary counting. By treating the two Cybersecurity workshops as one glued block, we arrange 5 items (3 identical AI, 1 Cybersecurity block, 1 Data Privacy), yielding 5!3!×1!×1!=20\frac{5!}{3! \times 1! \times 1!} = 20 forbidden arrangements. Subtracting the 20 forbidden arrangements from the 60 total arrangements gives 40 valid presentation schedules.

Adım Adım Çözüm

1
Calculate the total number of distinct schedules without restrictions.
Total arrangements = 6!3!×2!×1!=7206×2×1=60\frac{6!}{3! \times 2! \times 1!} = \frac{720}{6 \times 2 \times 1} = 60.
There are 66 total slots with 33 identical AI workshops, 22 identical Cybersecurity workshops, and 11 Data Privacy keynote.
2
Calculate the number of forbidden schedules where the 22 Cybersecurity workshops are in consecutive slots.
Forbidden arrangements = 5!3!×1!×1!=1206=20\frac{5!}{3! \times 1! \times 1!} = \frac{120}{6} = 20.
Treat the 22 identical Cybersecurity workshops as a single combined block. This leaves 55 items to arrange (33 AI, 11 combined Cybersecurity block, 11 Data Privacy).
3
Subtract the forbidden arrangements from the total arrangements using complementary counting.
Valid schedules = 60−20=4060 - 20 = 40.
The number of valid restricted arrangements is total arrangements minus restricted consecutive arrangements.

Anahtar Kavram

Counting with Restrictions and Repetitions (Complementary Counting and Permutations of Multisets)
Tahmini Süre:2m 0s
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Counting with Restrictions and Repetitions

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