To answer these questions, we will calculate the total number of possible strings and use the principle of combinations. Here are the key points:

1. There are 21 consonants and 5 vowels in the English alphabet.
2. Each string is composed of six lowercase letters.

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a) Exactly one vowel

1. Choose 1 vowel: $\binom{6}{1} = 6$ ways (choose which of the six positions will hold the vowel).
2. Choose the vowel: $5$ ways.
3. Fill the remaining 5 positions with consonants: $21^5$ ways.

The total number of strings is: $6 \times 5 \times 21^5$

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b) Exactly two vowels

1. Choose 2 positions for vowels: $\binom{6}{2} = 15$ ways.
2. Choose the vowels: $5^2 = 25$ ways.
3. Fill the remaining 4 positions with consonants: $21^4$ ways.

The total number of strings is: $15 \times 25 \times 21^4$

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c) At least one vowel

First, calculate the total number of strings and subtract the number of strings with no vowels:

• Total number of strings: $26^6$.
• Strings with no vowels (only consonants): $21^6$.

The total number of strings with at least one vowel is: $26^6 - 21^6$

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d) At least two vowels

We use complementary counting:

1. Total number of strings with at least one vowel: $26^6 - 21^6$.
2. Subtract the strings with exactly one vowel (calculated in part a).

The total number of strings with at least two vowels is: $(26^6 - 21^6) - \text{(result from part a)}$

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Next Steps

Let me know if you'd like me to calculate the numerical values for these expressions! Meanwhile:

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Relative Questions

1. How does the inclusion of uppercase letters affect the counting problem?
2. How would you approach this if the string length was variable (e.g., 3 to 6 letters)?
3. What if no letter can be repeated in the string? How does the counting change?
4. How would the problem change if a specific vowel must always appear?
5. What if we had a constraint that consonants and vowels alternate?

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Tip

When solving problems involving counting or probability, always start by defining the total possibilities and use complementary counting for "at least" or "at most" cases.