Probability
Permutations and probability
Grade 12
Question:
<p>Three-digit numbers are formed using the digits 0, 1, 2, 3, 4, 5 without repetition of digits. If a number is chosen at random, then the probability that the digits either increase or decrease, is</p>
<p>(a) \(\frac{1}{10}\)</p>
<p>(b) \(\frac{2}{11}\)</p>
<p>(c) \(\frac{3}{10}\)</p>
<p>(d) \(\frac{4}{11}\)</p>
Step-by-Step Solution
Key Concept: Count three-digit numbers from the given digits where digits form an increasing or decreasing sequence.
<p><strong>Solution:</strong> $n(S) = \text{Total number of three-digit numbers} = P(6,3) - P(5,2) = 120 - 20 = 100$</p><p>(We subtract cases where 0 is in the first position)</p><p>$n(E) = \text{Number of numbers with digits either increasing or decreasing} = 30$</p><p><strong>Required probability:</strong> $\frac{n(E)}{n(S)} = \frac{30}{100} = \frac{3}{10}$</p>
Correct Answer: C