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Question 3

menu_bookSolutionmenu_bookTheorymenu_book•(i)nPr=n⋅n−1Pr−1{}^nP_r = n \cdot {}^{n-1}P_{r-1}nPr​=n⋅n−1Pr−1​menu_book•(ii)nPr=n−1Pr+r⋅n−1Pr−1{}^nP_r = {}^{n-1}P_r + r \cdot {}^{n-1}P_{r-1}nPr​=n−1Pr​+r⋅n−1Pr−1​
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11TH · MATH

Question 3

menu_bookSolutionmenu_bookTheorymenu_book•(i)nPr=n⋅n−1Pr−1{}^nP_r = n \cdot {}^{n-1}P_{r-1}nPr​=n⋅n−1Pr−1​menu_book•(ii)nPr=n−1Pr+r⋅n−1Pr−1{}^nP_r = {}^{n-1}P_r + r \cdot {}^{n-1}P_{r-1}nPr​=n−1Pr​+r⋅n−1Pr−1​
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Accedmychevron_right11thchevron_rightmathchevron_rightPermutations And Combinationschevron_rightExercise 7.2

Proofs of the two identities from definition of ( {}^nP_r ).

Answer: Proved

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