(A) Given the equation: $^{n}P_{5} = 42 \, ^{n}P_{3}$.Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$n(n-1)(n-2)(n-3)(n-4) = 42 \, n(n-1)(

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(A) Given the equation: $^{n}P_{5} = 42 \, ^{n}P_{3}$.Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$n(n-1)(n-2)(n-3)(n-4) = 42 \, n(n-1)(n-2)$.Since $n > 4$,$n(n-1)(n-2) \neq 0$. Dividing both sides by $n(n-1)(n-2)$,we get:$(n-3)(n-4) = 42$.$n^{2} - 7n + 12 = 42$.$n^{2} - 7n - 30 = 0$.Factoring the quadratic equation:$(n-10)(n+3) = 0$.This gives $n = 10$ or $n = -3$.Since $n$ must be a positive integer and $n > 4$,we discard $n = -3$.Therefore,$n = 10$.

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

Medium

Find the value of $n$ such that $^{n}P_{5} = 42 \, ^{n}P_{3}$,where $n > 4$.

A
$10$
B
$11$
C
$12$
D
$9$

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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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Find the value of $n$ such that $^{n}P_{5} = 42 \, ^{n}P_{3}$,where $n > 4$.

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