समीकरण (x+y+z=12) के ऐसे पूर्णांक हल कितने हैं जिनमें \(x\ge2\), \(y\ge3\) और \(z\ge0\) हो?
How many integer solutions of (x+y+z=12) satisfy \(x\ge2\), \(y\ge3\), and \(z\ge0\)?
Explanation opens after your attempt
Correct Answer
A. (36)
Concept
After subtracting minimum values, (a+b+z=7), so \( \binom{9}{2}=36 \). First convert constraints to zero-based variables.
Why this answer is correct
The correct answer is A. (36). After subtracting minimum values, (a+b+z=7), so \( \binom{9}{2}=36 \). First convert constraints to zero-based variables.
Exam Tip
न्यूनतम मान घटाने पर (a+b+z=7) मिलता है, इसलिए \( \binom{9}{2}=36 \)। पहले शर्तों को शून्य-आधारित बनाइए।
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