Question

If \(N\) denotes number of 8-digit numbers that contain exactly four nines, then unit digit of \(N\) is:

Hint:

For digit problems, always subtract cases where the first digit is zero.

Answer 1

Correct Answer: 5

Concept: Count 8-digit numbers (first digit \(\neq 0\)) containing exactly four 9's.

Step 1: Total arrangements without leading zero restriction. Choose 4 positions for the four 9's: \[ \binom{8}{4} = 70 \] Remaining 4 positions can be filled with digits \(0\) to \(8\) (9 choices each): \[ 70 \times 9^4 \] This includes numbers starting with 0.

Step 2: Subtract numbers starting with 0. First digit fixed as 0 (not 9). From remaining 7 positions, choose 4 positions for 9's: \[ \binom{7}{4} = 35 \] Remaining 3 positions: digits \(0\) to \(8\) (9 choices each): \[ 35 \times 9^3 \]

Step 3: Valid count. \[ N = 70 \times 9^4 - 35 \times 9^3 \] \[ N = 35 \times 9^3 (2 \times 9 - 1) = 35 \times 9^3 \times 17 \]

Step 4: Find unit digit. \[ 9^1 \rightarrow 9,\quad 9^2 \rightarrow 1,\quad 9^3 \rightarrow 9 \] Unit digit of \(35 \times 17\): \[ 5 \times 7 = 35 \rightarrow \text{unit digit } 5 \] Unit digit of \(35 \times 9^3 \times 17\): \[ 5 \times 9 = 45 \rightarrow \text{unit digit } 5 \]