Question:

<p><strong>Step 1: Recognize the bijection</strong></p><p>A 6-digit number with digits in non-increasing order (e.g., 877550) is uniquely determined by choosing which digits appear and how many times. This is equivalent to selecting a 6-element multiset from {0,1,2,...,9}.</p><p><strong>Step 2: Count using stars and bars</strong></p><p>The number of such multisets is C(10+6-1, 6) = C(15, 6) = 5005. So there are 5005 valid numbers total, and the 2005th exists.</p><p><strong>Step 3: Find the 2005th by counting from largest digits</strong></p><p>Order by: numbers with first digit 9, then 8, etc. When listed in increasing order, we count backwards from the largest.</p><p>Numbers starting with 9: Choose 5 more from {0-9} = C(14,5) = 2002</p><p>Since 2005 > 2002, we need 2005 - 2002 = 3 more numbers starting with 8 or less.</p><p><strong>Step 4: Continue the decomposition</strong></p><p>Numbers starting with 8: Choose 5 from {0-8} = C(13,5) = 1287</p><p>The 3rd number overall comes from the (3rd) number in the sorted sequence with first digit ≤ 8.</p><p>Among numbers with digits from {0-8}: the 3rd smallest has first digit 8 and remaining 5 digits chosen to form the 3rd multiset from {0-8}.</p><p>Continuing this process: the largest 6 non-increasing numbers from {0-8} are found by choosing multisets in reverse lexicographic order.</p><p><strong>Step 5: Calculate precisely</strong></p><p>Working through the enumeration: the 2005th number is <strong>888888</strong> (taking the top 2002 numbers starting with 9, then the 3rd number overall in increasing order from what remains, which gives us 888888).</p><p>∴ Answer: A</p>