Question

Number of four digit numbers that can be formed using all the digits except zero such that every number has exactly \(2\) distinct digits in it is

• A. \(189\)
• B. \(216\)
• C. \(288\)
• D. \(504\)

Hint:

For numbers with exactly two distinct digits, first choose the two digits and then count all arrangements using both digits. For four places, this count is \(2^4-2\).

Answer 1

The Correct Option is D

Step 1: Identify the available digits.
We use digits $ 1, 2, 3, 4, 5, 6, 7, 8, 9 $ (zero is excluded). That gives us 9 digits total, all non-zero.
Step 2: Choose exactly 2 distinct digits.
We need each 4-digit number to use exactly 2 distinct digits. The number of ways to choose 2 digits from 9 is $ {}^9C_2 = \frac{9 \times 8}{2} = 36 $.
Step 3: Count arrangements using exactly 2 digits.
Once we pick digits $ a $ and $ b $, each of the 4 positions can be filled by either $ a $ or $ b $, giving $ 2^4 = 16 $ arrangements. But we must subtract the arrangements using only one digit ($ aaaa $ and $ bbbb $). So valid arrangements per pair: $ 16 - 2 = 14 $.
Step 4: Compute the total.
Total = $ 36 \times 14 = 504 $.
Step 5: Verify the logic.
The 14 arrangements ensure that both chosen digits appear at least once. The factor of 36 accounts for all pairs. No overcounting occurs because each pair gives a separate group of 14 numbers.
Step 6: State the final answer.
\[ \boxed{504} \]