# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-7/9

Exercises: Try Out - Set 7: Real exam simulation to sharpen your skills and build confidence. - Problem 9

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## Exercise 9

### Question

export const metadata = {
  title: "Problem 9",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/24/2025",
};

A code consists of <InlineMath math="2" /> vowels and <InlineMath math="2" /> digits in an alternating sequence, starting with a letter. The formation of the code follows these rules:

1. The first letter is not <InlineMath math="A" />.
2. The first digit is an odd number.
3. The second letter is the letter <InlineMath math="O" />.
4. The second digit is chosen from <InlineMath math="0, 1, 2, 4, \text{ and } 6" />.

How many codes can be formed if no letters or digits can be repeated?


### Choices

- [x] $$72$$
- [ ] $$96$$
- [ ] $$120$$
- [ ] $$144$$
- [ ] $$168$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 9",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/24/2025",
};

We need to form a code with the pattern **Letter - Digit - Letter - Digit**.

Let the slots for the code be:
<MathContainer>
  <BlockMath math="\underline{L_1} \times \underline{D_1} \times \underline{L_2} \times \underline{D_2}" />
</MathContainer>

Available sets:
- Vowels: <InlineMath math="\{A, I, U, E, O\}" /> (<InlineMath math="5" /> letters).
- Odd digits for <InlineMath math="D_1" />: <InlineMath math="\{1, 3, 5, 7, 9\}" /> (<InlineMath math="5" /> digits).
- Choice digits for <InlineMath math="D_2" />: <InlineMath math="\{0, 1, 2, 4, 6\}" /> (<InlineMath math="5" /> digits).

Rules and Constraints:
1. <InlineMath math="L_1 \neq A" />.
2. <InlineMath math="L_2 = O" />.
3. No repetition of letters or digits is allowed.

Since <InlineMath math="D_1" /> must be odd and <InlineMath math="D_2" /> is chosen from a specific set, there is an overlap at the digit <InlineMath math="1" />. We need to split the calculation into cases based on the value of <InlineMath math="D_2" />.

#### Case 1 If The Second Digit Is 1
If we choose the digit <InlineMath math="1" /> for position <InlineMath math="D_2" />:
1. <InlineMath math="L_2" /> (Second letter): Must be <InlineMath math="O" /> (<InlineMath math="1" /> way).
2. <InlineMath math="D_2" /> (Second digit): Must be <InlineMath math="1" /> (<InlineMath math="1" /> way).
3. <InlineMath math="L_1" /> (First letter): Vowel other than <InlineMath math="A" /> and other than <InlineMath math="O" /> (since <InlineMath math="O" /> is used in <InlineMath math="L_2" />).
   - Choices: <InlineMath math="\{I, U, E\}" /> (<InlineMath math="3" /> ways).
4. <InlineMath math="D_1" /> (First digit): Odd digit other than <InlineMath math="1" /> (since <InlineMath math="1" /> is used in <InlineMath math="D_2" />).
   - Choices: <InlineMath math="\{3, 5, 7, 9\}" /> (<InlineMath math="4" /> ways).

Total ways for Case <InlineMath math="1" />:
<MathContainer>
  <BlockMath math="3 \times 4 \times 1 \times 1 = 12 \text{ ways}" />
</MathContainer>

#### Case 2 If The Second Digit Is Not 1
If we choose a digit other than <InlineMath math="1" /> for position <InlineMath math="D_2" /> (which means it is even):
1. <InlineMath math="L_2" /> (Second letter): Must be <InlineMath math="O" /> (<InlineMath math="1" /> way).
2. <InlineMath math="D_2" /> (Second digit): Chosen from <InlineMath math="\{0, 2, 4, 6\}" /> (<InlineMath math="4" /> ways).
3. <InlineMath math="L_1" /> (First letter): Vowel other than <InlineMath math="A" /> and other than <InlineMath math="O" />.
   - Choices: <InlineMath math="\{I, U, E\}" /> (<InlineMath math="3" /> ways).
4. <InlineMath math="D_1" /> (First digit): Odd digit <InlineMath math="\{1, 3, 5, 7, 9\}" />. Since <InlineMath math="D_2" /> is even, there is no conflict. All odd digits can be chosen.
   - Choices: <InlineMath math="5" /> ways.

Total ways for Case <InlineMath math="2" />:
<MathContainer>
  <BlockMath math="3 \times 5 \times 1 \times 4 = 60 \text{ ways}" />
</MathContainer>

#### Total Combinations
Sum the results from both cases:
<MathContainer>
  <BlockMath math="12 + 60 = 72" />
</MathContainer>

Therefore, the number of codes that can be formed is <InlineMath math="72" />.


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