Step-by-step Solution:

Assertion \( A \): An elevator starts with \( m \) passengers and stops at \( n \) floors (\( m \leq n \)). The probability that no passengers alight at the same floor is: \[ \frac{n_P m}{m^m} \] Derivation of Probability Each passenger can choose any of the \( n \) floors independently. The total number of ways to assign \( m \) passengers to \( n \) floors is: \[ n^m \] If no two passengers choose the same floor, then this is equivalent to arranging \( m \) passengers into \( n \) floors uniquely. The number of ways to do this is given by a permutation: \[ P(n, m) = \frac{n!}{(n-m)!} \] Thus, the required probability is: \[ \frac{P(n, m)}{n^m} = \frac{n!}{(n-m)! n^m} \] Since this formula matches Assertion \( A \), we conclude that Assertion \( A \) is correct. --- Reason \( R \): If \( (n+1) \) is an integer, say \( m \), then the probability mass function of the binomial distribution: \[ P(X = r) = \binom{n}{r} p^r (1 - p)^{n-r} \] is maximized when \( r = m \) or \( r = m-1 \). Explanation - In a binomial distribution \( P(X = r) \), the mode of the distribution (the value of \( r \) where the probability is highest) is given by: \[ r = \lfloor (n+1) p \rfloor \quad \text{or} \quad r = \lfloor (n+1) p \rfloor - 1 \] This is a well-known result for the mode of the binomial distribution. Thus, Reason \( R \) is also correct. --- Checking Whether \( R \) Explains \( A \): - Assertion \( A \) deals with uniform random assignment of passengers to floors, leading to a probability formula derived using combinatorial counting. - Reason \( R \) describes the mode of a binomial distribution, which is unrelated to the way passengers randomly choose floors. Since \( R \) does not explain \( A \), but both statements are independently correct, the correct answer is: \[ \textbf{B. Both A and R are correct but R is not the correct explanation of A} \]