Examkin

Soru

Zorluk: OrtaCounting with Restrictions and Repetitions

A 6-letter security code is created by arranging all the letters in the word BANANABANANA. How many distinct security codes can be formed such that the two letters NN are not adjacent to each other?

  1. A
    20
  2. B
    30
  3. 40Cevap
  4. D
    60
  5. E
    120

Cevap

40 distinct security codes can be formed such that the two NN's are not adjacent.
To find the number of arrangements where the two NN's are not adjacent, use complementary counting. First, compute total unrestricted permutations of BANANABANANA (1 BB, 3 AA's, 2 NN's), which gives 6!3!×2!=60\frac{6!}{3! \times 2!} = 60. Next, count the forbidden arrangements where the two NN's are together by grouping them into a single block (NN)(NN). Arranging BB, AA, AA, AA, and (NN)(NN) yields 5!3!=20\frac{5!}{3!} = 20. Subtracting forbidden arrangements from total arrangements gives 60−20=4060 - 20 = 40.

Adım Adım Çözüm

1
Calculate the total number of distinct arrangements of the letters in BANANABANANA without restrictions.
The word BANANABANANA contains 6 letters in total: 1 BB, 3 AA's, and 2 NN's. The number of distinct permutations is 6!1!×3!×2!=7201×6×2=60\frac{6!}{1! \times 3! \times 2!} = \frac{720}{1 \times 6 \times 2} = 60.
Indistinguishable duplicate letters must be accounted for by dividing the total permutations by the factorial of each letter's count.
2
Calculate the number of arrangements where the two NN's ARE adjacent.
Treat the pair NNNN as a single block. The items to arrange are BB, AA, AA, AA, and (NN)(NN), which gives 5 items with 3 identical AA's. The number of arrangements is 5!1!×3!×1!=1206=20\frac{5!}{1! \times 3! \times 1!} = \frac{120}{6} = 20.
Grouping restricted elements into a single composite element simplifies calculating forbidden configurations.
3
Apply complementary counting to find the number of valid arrangements where the two NN's are not adjacent.
Total valid arrangements = Total arrangements - Adjacent arrangements = 60−20=4060 - 20 = 40.
Complementary counting is the most efficient method when calculating non-adjacent placements.

Anahtar Kavram

Counting with Restrictions and Repetitions (Complementary Counting Principle)
Tahmini Süre:1m 30s
Bu soruyu puanla

Konular

Counting with Restrictions and Repetitions

Bu konudan daha fazla soru