(D) Given that ${}^n P_4 = 1680$. Using the formula ${}^n P_r = \frac{n!}{(n-r)!}$,we have: $n(n-1)(n-2)(n-3) = 1680$. We need to find four consecutiv

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(D) Given that ${}^n P_4 = 1680$. Using the formula ${}^n P_r = \frac{n!}{(n-r)!}$,we have: $n(n-1)(n-2)(n-3) = 1680$. We need to find four consecutive integers whose product is $1680$. By prime factorization: $1680 = 8 \times 7 \times 6 \times 5$. Comparing the terms,we get $n = 8$. Hence,option $D$ is correct.

Class 11 Mathematics Permutations and Combinations Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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If ${}^n P_4 = 1680$,then $n =$

AP EAMCET 2020 All AP EAMCET

A
$6$
B
$12$
C
$10$
D
$8$

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Permutations and Combinations

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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If ${}^n P_4 = 1680$,then $n =$

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