WBJEE · Maths · Permutation Combination
Let \(x_{1}, x_{2}, \ldots, x_{13}\) be 15 distinct numbers chosen from \(1,2,3, \ldots, 15 .\) Then, the value of \(\left(x_{1}-1\right)\left(x_{2}-1\right)\left(x_{3}-1\right) \ldots\left(x_{15}-1\right)\) is
- A always \(\leq 0\)
- B 0
- C akays even
- D aways odd
Answer & Solution
Correct Answer
(B) 0
Step-by-step Solution
Detailed explanation
Since, \(x_{1}, x_{2} \ldots, x_{15}\) are among \(1,2,3, \ldots, 15\). So, \(x_{1}\) can take any value from \(1,2,3, \ldots, 15\). Among these values, one of the number must be 1, hence product will be 0
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