Suppose r integers, 0 < r < 10, are chosen from (0, 1, 2, ...,9) at random and with replacement. The probability that no two are equal, is
Step-by-step Solution:
Given: \( r \) integers, \( 0 < r < 10 \), are chosen from \( \{0, 1, 2, \dots, 9\} \) at random and with replacement. 1. Total Possible Outcomes: \[ 10^r \] 2. Favorable Outcomes (No Two Integers Are Equal): \[ \binom{10}{r} \] 3. Probability that No Two Integers Are Equal: \[ \frac{\binom{10}{r}}{10^r} \] Expanding the combination formula: \[ \frac{\frac{10!}{r!(10 - r)!}}{10^r} \] \[ = \frac{10!}{10^r \times r! \times (10 - r)!} \]