(n) letters की derangement संख्या \(D_n\) के लिए recurrence (D_n=(n-1)\(D_{n-1}+D_{n-2}\)) में (n-1) factor क्यों आता है?
In the derangement recurrence (D_n=(n-1)\(D_{n-1}+D_{n-2}\)), why does the factor (n-1) appear?
Explanation opens after your attempt
A. पहले letter की गलत position चुनने के लिएTo choose the wrong position of the first letter
Concept
The first letter can go to (n-1) positions other than its original position. In exams watch the first object's wrong choice in derangement recurrence.
Why this answer is correct
The correct answer is A. पहले letter की गलत position चुनने के लिए / To choose the wrong position of the first letter. The first letter can go to (n-1) positions other than its original position. In exams watch the first object's wrong choice in derangement recurrence.
Exam Tip
पहला letter अपनी original position छोड़कर (n-1) positions में जा सकता है। परीक्षा में derangement recurrence में first object की गलत choice देखें।
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