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۱۱) الف) با حروف کلمه \البرز\ چند کلمه 5 حرفی می توان نوشت که در آن دو …

Question

۱۱) الف) با حروف کلمه \البرز\ چند کلمه 5 حرفی می توان نوشت که در آن دو حرف \ر\ و \ز\ در کنار هم آمده باشند؟

ب) با حروف کلمه \البرز\ چند کلمه 4 حرفی می توان نوشت که در آن دو حرف \ر\ و \ز\ در کنار هم آمده باشند؟

Explanation:

🆕 New Concept Discovered: Permutations with Grouping Constraints
Grouping items to keep them together in arrangements.

Step 1: Analyze the letters of the word "البرز"

The word "البرز" (Alborz) consists of 5 distinct Persian letters:

$$ \{\text{ا}, \text{ل}, \text{ب}, \text{ر}, \text{ز}\} $$

Let's represent them as:

  • \( \text{ا} \) (A)
  • \( \text{ل} \) (L)
  • \( \text{ب} \) (B)
  • \( \text{ر} \) (R)
  • \( \text{ز} \) (Z)

We need to find arrangements where the letters "ر" (R) and "ز" (Z) are adjacent (next to each other).

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Step 2: Solve Part A (5-letter words)

To find the number of 5-letter words where "ر" and "ز" are adjacent:

  1. Treat the block \( (\text{ر}, \text{ز}) \) as a single super-letter.
  2. This leaves us with 4 items to arrange: \( \{\text{ا}\}, \{\text{ل}\}, \{\text{ب}\}, \{\text{ر}, \text{ز}\} \).
  3. The number of ways to arrange these 4 items is:
$$ 4! = 24 $$
  1. Within the block, the letters "ر" and "ز" can switch places in \( 2! \) ways:
$$ 2! = 2 $$
  1. Total arrangements for Part A:
$$ 4! \times 2! = 24 \times 2 = 48 $$

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Step 3: Solve Part B (4-letter words)

To find the number of 4-letter words where "ر" and "ز" are adjacent:

  1. Since "ر" and "ز" must be in the word and adjacent, we treat them as a single block \( (\text{ر}, \text{ز}) \).
  2. We need to choose 2 more letters from the remaining 3 letters \( \{\text{ا}, \text{ل}, \text{ب}\} \) to complete the 4-letter word.
  3. The number of ways to choose 2 letters from 3 is:
$$ \binom{3}{2} = 3 $$
  1. Now we have 3 items to arrange: the block \( (\text{ر}, \text{ز}) \) and the 2 chosen letters. The number of ways to arrange these 3 items is:
$$ 3! = 6 $$
  1. Within the block, "ر" and "ز" can switch places in \( 2! \) ways:
$$ 2! = 2 $$
  1. Total arrangements for Part B:
$$ \binom{3}{2} \times 3! \times 2! = 3 \times 6 \times 2 = 36 $$

Answer:

  • Part A (الف): \( 48 \)
  • Part B (ب): \( 36 \)