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Zorluk: Çok zorPermutations, Combinations, and Fundamental Counting Principle

An academic conference has 7 consecutive presentation time slots. A committee must assign 7 presentations—3 in Biology (BB), 2 in Chemistry (CC), and 2 in Physics (PP)—to these slots subject to the following conditions:

1. Presentations of the same discipline are indistinguishable (only the subject sequence matters).
2. No two Biology presentations may be scheduled in consecutive time slots.
3. The two Physics presentations must be scheduled in consecutive time slots.

How many different subject-sequence schedules for the 7 presentation time slots satisfy all of these conditions?

  1. A
    2424
  2. 1212Cevap
  3. C
    3030
  4. D
    6060
  5. E
    7272

Cevap

12
To satisfy the condition that the two Physics presentations are consecutive, we combine them into a single block (PP)(PP). We then arrange the non-Biology items—two Chemistry presentations CC and one block (PP)(PP)—which can be ordered in 3!2!=3\frac{3!}{2!} = 3 distinct ways. These 3 items create 4 potential gaps (before the first item, between items, and after the last item). To ensure no two Biology presentations are adjacent, we place one Biology presentation into each of 3 chosen gaps out of the 4 available spaces, which can be done in (43)=4\binom{4}{3} = 4 ways. By the Fundamental Counting Principle, the total number of valid subject-sequence schedules is 3×4=123 \times 4 = 12.

Adım Adım Çözüm

1
Group the adjacent Physics presentations into a single block
The two Physics presentations (P,P)(P, P) form a single block (PP)(PP). The set of non-Biology items consists of two indistinguishable CC's and one (PP)(PP) block, making 3 items in total.
Since the two Physics presentations must be scheduled in consecutive slots, treating them as a single unit guarantees they stay adjacent.
2
Calculate the number of distinct arrangements of the non-Biology items
The number of distinct arrangements of {C,C,(PP)}\{C, C, (PP)\} is 3!2!1!=3\frac{3!}{2!1!} = 3 ways.
The 2 Chemistry presentations are identical, so we divide the total permutations (3!3!) by 2!2! to account for indistinguishability.
3
Apply the gap method to place the non-adjacent Biology presentations
Any sequence of the 3 non-Biology items creates 4 available spaces (gaps), including the two ends: _item1_item2_item3_\_ \text{item}_1 \_ \text{item}_2 \_ \text{item}_3 \_. Choosing 3 gaps out of 4 yields (43)=4\binom{4}{3} = 4 ways to insert the 3 identical BB presentations.
Placing at most one Biology presentation in each gap guarantees no two Biology presentations are consecutive.
4
Apply the Fundamental Counting Principle
Total valid schedules = 3×4=123 \times 4 = 12.
The choices of arranging non-Biology items and selecting gaps for Biology items are independent.

Anahtar Kavram

Combinatorics with Adjacency Restrictions and Indistinguishable Objects
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Konular

Permutations, Combinations, and Fundamental Counting Principle

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