Let \( \mathrm{n}_{1}\mathrm{n}_{2}\mathrm{n}_{3}\mathrm{n}_{4}\mathrm{n}_{5} \) be positive int...

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Let \( \mathrm{n}_{1}\mathrm{n}_{2}\mathrm{n}_{3}\mathrm{n}_{4}\mathrm{n}_{5} \) be positive integers such that \( \mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}+\mathrm{n}_{4}+\mathrm{n}_{5}=20 \). Then the number of such distinct arrangements \( \left(\mathrm{n}_{1}, \mathrm{n}_{2}, \mathrm{n}_{3}, \mathrm{n}_{4}, \mathrm{n}_{5}\right) \) is
[JEE (Advanced) 2014, 3]
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Supper explanation for where you saw the solution sir 😂😂😂

mastermadan