<p>If <em>n</em><sub>1</sub> and <em>n</em><sub>2</sub> are five-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 these numbers can be added without carrying at any stage.</p>
Step-by-Step Solution
Key Concept: For no carrying to occur when adding two five-digit numbers digit by digit, the sum of corresponding digits must be ≤ 9. Each digit position has independent constraints: if first number has digit a and second has digit b, then a + b ≤ 9.
<p><strong>Step 1:</strong> For a five-digit number, the first digit must be from {1,2,...,9} and remaining digits from {0,1,...,9}.</p><p><strong>Step 2:</strong> For the ten-thousands place (first digit): If n₁ has digit a₁ and n₂ has digit b₁, both from {1,2,...,9}, we need a₁ + b₁ ≤ 9. Pairs satisfying this: (1,1), (1,2),...,(1,8), (2,1),...,(2,7),...,(8,1). Count = 8+7+6+5+4+3+2+1 = 36 pairs.</p><p><strong>Step 3:</strong> For each of the remaining 4 digit positions (thousands, hundreds, tens, units): If n₁ has digit aᵢ ∈ {0,1,...,9} and n₂ has digit bᵢ ∈ {0,1,...,9}, we need aᵢ + bᵢ ≤ 9. Pairs: (0,0), (0,1),...,(0,9), (1,0),...,(1,8),...,(9,0). Count = 10+9+8+7+6+5+4+3+2+1 = 55 pairs.</p><p><strong>Step 4:</strong> By multiplication principle, total ways = 36 × 55 × 55 × 55 × 55 = 36(55)⁴.</p><p>∴ Answer: <strong>36(55)⁴</strong></p>
Correct Answer: 36(55)^4