<p>For three digit numbers formed using digits from a set of 5 digits (where 0 is not included), and four digit numbers similarly formed, find the total count of required numbers.</p>
Step-by-Step Solution
Key Concept: For three-digit numbers, all 5 digits can be used in any position since 0 is excluded. For four-digit numbers, we select and arrange 4 digits from 5 available digits. The total is the sum of both cases.
<p><strong>Step 1: Count three-digit numbers</strong></p><p>Since 0 is not in the set, we have 5 digits available with no restrictions on any position.</p><p>• First digit: 5 choices</p><p>• Second digit: 5 choices</p><p>• Third digit: 5 choices</p><p>Three-digit numbers = 5 × 5 × 5 = 125</p><p><strong>Step 2: Count four-digit numbers</strong></p><p>We need to select and arrange 4 digits from 5 available digits (without repetition).</p><p>• First digit: 5 choices</p><p>• Second digit: 4 choices (one already used)</p><p>• Third digit: 3 choices</p><p>• Fourth digit: 2 choices</p><p>Four-digit numbers = P(5,4) = 5 × 4 × 3 × 2 = 120</p><p><strong>Step 3: Find total</strong></p><p>Alternatively: P(5,4) = 5!/(5-4)! = 5!/1! = 120</p><p>Total required numbers = 125 + 120 = 245</p><p><em>Note: If answer is 374, the problem likely includes additional cases (like numbers with repetition allowed for four-digit, or uses 6 digits instead of 5). With standard interpretation: 5³ + P(5,4) = 125 + 120 = 245. Please verify problem statement for 374 result.</em></p>
Correct Answer: 374