 How many positive integers n can be formed using the digits 3, 4, 5, 6, 6, 7,7, 
8, if n has to exceed 60,00,000?

We are given a set of digits: \(3, 4, 5, 6, 6, 7, 7, 8\) and we are tasked with finding how many different integers \(n\) can be formed such that \(n > 60,00,000\) (6 million).

### Step 1: Number of digits
Since the number \(n\) must exceed 6 million, \(n\) must be at least 7 digits long. Therefore, \(n\) will be a 7-digit number or an 8-digit number.

### Step 2: Counting the 7-digit numbers
To form a 7-digit number, we must select and arrange 7 digits from the set of available digits.

#### Case 1: First digit is 6, 7, or 8
For \(n > 6,000,000\), the first digit must be either 6, 7, or 8. Let's consider each case:

- **First digit is 6**: 
  We are left with the digits \(3, 4, 5, 6, 7, 7, 8\). These 7 digits include a repetition of the digit 7, so the number of distinct arrangements of these digits is:
  \[
  \frac{7!}{2!} = \frac{5040}{2} = 2520
  \]
  
- **First digit is 7**:
  We are left with the digits \(3, 4, 5, 6, 6, 7, 8\). These 7 digits include a repetition of the digit 6, so the number of distinct arrangements is:
  \[
  \frac{7!}{2!} = \frac{5040}{2} = 2520
  \]

- **First digit is 8**:
  We are left with the digits \(3, 4, 5, 6, 6, 7, 7\). These 7 digits include repetitions of both 6 and 7, so the number of distinct arrangements is:
  \[
  \frac{7!}{2!2!} = \frac{5040}{4} = 1260
  \]

The total number of 7-digit numbers is:
\[
2520 + 2520 + 1260 = 6300
\]

### Step 3: Counting the 8-digit numbers
All 8 digits can be used to form an 8-digit number. The number of distinct arrangements, considering the two 6's and the two 7's, is:
\[
\frac{8!}{2!2!} = \frac{40320}{4} = 10080
\]

### Step 4: Total number of valid numbers
The total number of integers \(n\) that exceed 6 million is the sum of the 7-digit and 8-digit numbers:
\[
6300 + 10080 = 16380
\]

Thus, the total number of positive integers \(n\) that can be formed using the digits \(3, 4, 5, 6, 6, 7, 7, 8\) and exceed 60,00,000 is **16,380**.

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Do you want details on any step or have any questions?

Here are 5 related questions:
1. How would the count change if repetitions of 6 and 7 were not allowed?
2. What if \(n\) had to be less than 60,00,000 instead of exceeding it?
3. How many 7-digit numbers can be formed if only even digits are allowed in the first place?
4. How does the formula change if there were 3 repeated digits instead of 2?
5. What if the first digit could not be 6, 7, or 8?

**Tip**: When arranging objects with repeated elements, always account for these repetitions by dividing by the factorial of the number of repeated objects.

How many positive integers n can be formed using the digits 3, 4, 5, 6, 6, 7, 7, 8, if n has to exceed 60,00,000?

Find out how many integers greater than 60,00,000 can be formed using the digits 3, 4, 5, 6, 6, 7, 7, 8. This combinatorics problem involves calculating permutations with repeated digits and is suitable for Grades 10-12.

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