Let x_{1}, x_{2}, ldots, x_{15} be 15 distinc | vedclass

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(B) Given that $x_{1}, x_{2}, \ldots, x_{15}$ are $15$ distinct numbers chosen from the set $\{1, 2, 3, \ldots, 15\}$.Since there are exactly $15$ dis

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(B) Given that $x_{1}, x_{2}, \ldots, x_{15}$ are $15$ distinct numbers chosen from the set $\{1, 2, 3, \ldots, 15\}$.Since there are exactly $15$ distinct numbers in the set and we are choosing $15$ distinct numbers,the set $\{x_{1}, x_{2}, \ldots, x_{15}\}$ must be exactly the set $\{1, 2, 3, \ldots, 15\}$ in some order.Therefore,one of the values $x_{i}$ must be equal to $1$.If $x_{i} = 1$ for some $i \in \{1, 2, \ldots, 15\}$,then the term $(x_{i}-1) = (1-1) = 0$.Since the product contains a factor of $0$,the entire product $(x_{1}-1)(x_{2}-1) \ldots (x_{15}-1)$ is equal to $0$.

Let $x_{1}, x_{2}, \ldots, x_{15}$ be $15$ distinct numbers chosen from $1, 2, 3, \ldots, 15$. Then,the value of $(x_{1}-1)(x_{2}-1)(x_{3}-1) \ldots (x_{15}-1)$ is

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