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Zorluk: ZorCounting with Restrictions and Repetitions

An executive board must arrange 7 books side-by-side on a display shelf: 3 identical copies of a Finance textbook, 2 identical copies of an Economics textbook, and 2 distinct novels (Novel A and Novel B). How many distinct linear arrangements of these 7 books are possible such that Novel A and Novel B are not placed adjacent to each other?

Cevap: 300

Cevap

300
To find the number of distinct arrangements where Novel A and Novel B are not adjacent, use complementary counting. First, compute the total distinct arrangements of the 7 books with identical items: 7! / (3! * 2!) = 420. Next, find the number of arrangements where Novel A and Novel B ARE adjacent by bundling them as one unit [AB], which has 2 internal orderings (AB or BA). Arranging the 6 units ([AB], 3 Finance, 2 Economics) gives 6! / (3! * 2!) = 60 ways. Multiplying by 2 internal orderings gives 120 adjacent arrangements. Subtracting these from the total yields 420 - 120 = 300.

Adım Adım Çözüm

1
Calculate the unrestricted total number of distinct permutations of the 7 books.
420 distinct arrangements
There are 7 total items with 3 identical Finance books and 2 identical Economics books. Using the formula N! / (n1! * n2! ...), we get 7! / (3! * 2!) = 5,040 / 12 = 420.
2
Calculate the number of forbidden arrangements where Novel A and Novel B are adjacent.
120 forbidden arrangements
Treat Novel A and Novel B as a single composite item [AB]. There are 2! = 2 ways to order A and B inside this block (AB or BA). Then, arrange the 6 items ([AB] and the 5 textbooks with repetitions): 6! / (3! * 2!) = 60. Multiplying by 2 internal orderings gives 2 * 60 = 120.
3
Subtract the forbidden (adjacent) arrangements from the total unrestricted arrangements.
300 valid arrangements
By complementary counting, Valid = Total - Forbidden = 420 - 120 = 300.

Anahtar Kavram

Counting permutations with indistinguishable items and non-adjacency restrictions using complementary counting
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