(\sum_{r=0}^{n}r(r-1){}^{n}C_r) का सही सरल रूप कौन-सा है?

What is the correct simplified form of (\sum_{r=0}^{n}r(r-1){}^{n}C_r)?

Explanation opens after your attempt
Correct Answer

B. (n(n-1)2^{n-2})

Step 1

Concept

There are (n(n-1)) ways to choose two ordered marked members and the remaining (n-2) are chosen freely. In exams two marks lead to \(2^{n-2}\).

Step 2

Why this answer is correct

The correct answer is B. (n(n-1)2^{n-2}). There are (n(n-1)) ways to choose two ordered marked members and the remaining (n-2) are chosen freely. In exams two marks lead to \(2^{n-2}\).

Step 3

Exam Tip

Ordered दो marked members चुनने के (n(n-1)) ways हैं और बाकी (n-2) freely चुने जाते हैं। परीक्षा में दो marks हों तो \(2^{n-2}\) आता है।

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Mathematics Answer, Explanation and Revision Hints

(\sum_{r=0}^{n}r(r-1){}^{n}C_r) का सही सरल रूप कौन-सा है? / What is the correct simplified form of (\sum_{r=0}^{n}r(r-1){}^{n}C_r)?

Correct Answer: B. (n(n-1)2^{n-2}). Explanation: Ordered दो marked members चुनने के (n(n-1)) ways हैं और बाकी (n-2) freely चुने जाते हैं। परीक्षा में दो marks हों तो \(2^{n-2}\) आता है। / There are (n(n-1)) ways to choose two ordered marked members and the remaining (n-2) are chosen freely. In exams two marks lead to \(2^{n-2}\).

Which concept should I revise for this Mathematics MCQ?

There are (n(n-1)) ways to choose two ordered marked members and the remaining (n-2) are chosen freely. In exams two marks lead to \(2^{n-2}\).

What exam hint can help solve this Mathematics question?

Ordered दो marked members चुनने के (n(n-1)) ways हैं और बाकी (n-2) freely चुने जाते हैं। परीक्षा में दो marks हों तो \(2^{n-2}\) आता है।