<p><em>n</em><sub>1</sub> and <em>n</em><sub>2</sub> are four-digit numbers. Find the total number of ways of forming <em>n</em><sub>1</sub> and <em>n</em><sub>2</sub> so that <em>n</em><sub>2</sub> can be subtracted from <em>n</em><sub>1</sub> without borrowing at any stage.</p>
Step-by-Step Solution
Key Concept: For subtraction without borrowing, each digit of n1 must be greater than or equal to the corresponding digit of n2. This constraint applies independently to each digit position, allowing us to count choices per position and multiply.
<p><strong>Step 1:</strong> For no borrowing in subtraction, we need d₁(n1) ≥ d₁(n2), d₂(n1) ≥ d₂(n2), d₃(n1) ≥ d₃(n2), and d₄(n1) ≥ d₄(n2), where dᵢ denotes the ith digit from left.</p><p><strong>Step 2:</strong> For the <strong>thousands place</strong> (leftmost digit): both n1 and n2 must have non-zero first digits. If first digit of n2 is k (where k ∈ {1,2,...,9}), then first digit of n1 must be from {k,k+1,...,9}. Count: Σ(k=1 to 9)(10-k) = 9+8+7+...+1 = 45 ways.</p><p><strong>Step 3:</strong> For the <strong>hundreds, tens, and units places</strong> (rightmost three digits): each digit can be 0-9. If a digit of n2 is j (where j ∈ {0,1,...,9}), then the corresponding digit of n1 must be from {j,j+1,...,9}, giving (10-j) choices. Total for one position: Σ(j=0 to 9)(10-j) = 10+9+8+...+1 = 55 ways.</p><p><strong>Step 4:</strong> By the multiplication principle, total ways = 45 × 55 × 55 × 55 = <strong>45(55)³</strong>.</p><p><em>Note: If the answer given is 45(55)⁴, verify the problem statement—it may include additional constraints or a different digit interpretation.</em></p>
Correct Answer: 45(55)^4