the letter without repetitions,how many 4 letter codes can be formed,if it contain 3 consonants

To solve the problem of how many 4-letter codes can be formed using 3 consonants and 1 vowel with no repetitions, we will break it down step by step.

### Step 1: Determine the number of consonants and vowels.
- There are 21 consonants in the English alphabet.
- There are 5 vowels in the English alphabet.

### Step 2: Choose the letters.
- **Choosing 3 consonants:** The number of ways to choose 3 consonants out of 21 is given by \( \binom{21}{3} = \frac{21!}{3!(21-3)!} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1} = 1330 \).
- **Choosing 1 vowel:** The number of ways to choose 1 vowel out of 5 is given by \( \binom{5}{1} = 5 \).

### Step 3: Arrange the letters.
- The 4 chosen letters (3 consonants + 1 vowel) can be arranged in \( 4! \) (4 factorial) ways. \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).

### Step 4: Calculate the total number of codes.
The total number of codes is the product of the number of ways to choose the letters and the number of ways to arrange them:

\[
\text{Total number of codes} = \binom{21}{3} \times \binom{5}{1} \times 4!
\]

\[
\text{Total number of codes} = 1330 \times 5 \times 24 = 159,600
\]

### Final Answer:
There are **159,600** possible 4-letter codes that can be formed using 3 consonants and 1 vowel, with no repetitions.

Would you like more details or have any questions?

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Here are some related questions you might find interesting:

1. How many 4-letter codes can be formed using only vowels?
2. What if the codes had to be 5 letters long instead of 4?
3. How many 4-letter codes can be formed using 2 consonants and 2 vowels?
4. What if repetitions were allowed in forming the 4-letter codes?
5. How many 4-letter codes can be formed using exactly 2 consonants?

**Tip:** Factorials grow very quickly, so breaking down a problem into steps and calculating each part separately can help avoid mistakes.

Discover how many 4-letter codes can be formed using 3 consonants and 1 vowel with no repetitions. Follow a detailed mathematical solution involving combinatorics and permutations. Suitable for students in Grades 10-12.

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