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Zorluk: OrtaPermutations and Linear Arrangements

A secure access code is formed by arranging all 66 of the distinct characters A,E,B,C,D,A, E, B, C, D, and FF in a single line. If the two vowels (AA and EE) cannot be adjacent to each other, and the code cannot begin with the consonant BB, how many different valid access codes can be formed?

  1. A
    240
  2. B
    252
  3. C
    360
  4. 408Cevap
  5. E
    480

Cevap

408 valid access codes can be formed.
To find the number of valid arrangements satisfying both conditions, we use complementary counting. First, the total number of arrangements of 6 distinct characters is 6! = 720. Next, we determine the number of invalid arrangements. Let V be the set of arrangements where A and E are adjacent: treating (AE) as a single unit gives 5! × 2! = 240 ways. Let B_first be the set of arrangements starting with B: fixing B in the first spot leaves 5! = 120 ways. The intersection of these two invalid sets (where A and E are adjacent AND the code starts with B) places B in position 1 and arranges the remaining 4 items including the (AE) block in 4! × 2! = 48 ways. By inclusion-exclusion, the total invalid arrangements equal 240 + 120 - 48 = 312. Subtracting invalid arrangements from the total gives 720 - 312 = 408 valid access codes.

Adım Adım Çözüm

1
Calculate the total total number of unrestricted linear arrangements of the 6 distinct characters.
6! = 720 total arrangements.
6 distinct characters can be ordered in 6! different ways.
2
Calculate the number of arrangements where the two vowels (A and E) ARE adjacent (Set V).
|V| = 240 arrangements.
Treating (AE) as 1 block leaves 5 items to arrange in 5! = 120 ways. Since (AE) has 2! = 2 internal arrangements, |V| = 2 * 120 = 240.
3
Calculate the number of arrangements that BEGIN with the consonant B (Set B).
|B| = 120 arrangements.
Fixing B in the first position leaves 5 remaining positions to be filled by the remaining 5 characters in 5! = 120 ways.
4
Calculate the overlap: arrangements where vowels ARE adjacent AND the code BEGINS with B (Set V ∩ B).
|V ∩ B| = 48 arrangements.
With B fixed in position 1, the remaining 5 positions are filled by 4 blocks: (AE), C, D, F. These 4 blocks can be arranged in 4! = 24 ways, and (AE) has 2! = 2 internal orderings, giving 24 * 2 = 48.
5
Apply the Principle of Inclusion-Exclusion to find the total invalid arrangements, and subtract from the total.
720 - (240 + 120 - 48) = 720 - 312 = 408 valid arrangements.
The number of invalid arrangements is |V ∪ B| = |V| + |B| - |V ∩ B| = 312. Subtracting invalid codes from total codes yields 408.

Anahtar Kavram

Linear permutations with multiple restrictions using complementary counting and inclusion-exclusion.
Tahmini Süre:2m 0s
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Konular

Permutations and Linear Arrangements

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