Step 1: Understanding the Concept:
We need to count 8-digit numbers with exactly four 9s. We split the problem into two cases based on whether the first digit is 9 or not. Step 2: Detailed Explanation: Case 1: First digit is 9.
- First spot is fixed (1 way).
- We need 3 more 9s from the remaining 7 positions: \(\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\) ways.
- The remaining 4 positions can be any of the other 9 digits (\(0-8\)): \(9^{4}\) ways.
- Total for Case 1 \(= 35 \times 9^{4}\). Case 2: First digit is NOT 9.
- First spot must be one of the digits \(\{1, 2, \dots, 8\}\): 8 ways.
- We need exactly four 9s from the remaining 7 positions: \(\binom{7}{4} = 35\) ways.
- The remaining 3 positions can be any of the other 9 digits (\(0-8\)): \(9^{3}\) ways.
- Total for Case 2 \(= 8 \times 35 \times 9^{3}\).
Summing both cases:
\[ N = (35 \times 9^{4}) + (8 \times 35 \times 9^{3}) = 35 \times 9^{3} \times (9 + 8) = 35 \times 729 \times 17 \]
To find the unit digit of \(N\), find the product of unit digits:
\[ \text{Unit digit of } N = \text{Unit digit of } (5 \times 9 \times 7) = \text{Unit digit of } 315 = 5 \] Step 3: Final Answer:
The unit digit of \(N\) is 5.