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Permutations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsPermutations3 Marks
Question
If $(n + 3)! = 56 [(n + 1)!],$ find $n.$
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Answer

We have,$(n + 3)! = 56 [(n + 1)!]$
$\Rightarrow (n + 3)! \times (n + 2)! \times (n + 1)! = 56 [(n - 1)!] $
$\Rightarrow (n + 2)(n + 3) = 56 $
$\Rightarrow n^2 + 3n + 2n + 6 = 56$
$\Rightarrow n^2 + 5n + 6 - 56 = 0 $
$\Rightarrow n^2 + 5n + 50 = 0 $
$\Rightarrow n^2 + 10n - 5n - 50 = 0 $
$\Rightarrow n (n + 10) - 5 (n + 10) = 0 $
$\Rightarrow (n - 5)(n + 10) = 0$
$\big[\therefore \text{n} +10 \neq 0\big]$
$\Rightarrow n - 5 = 0$
$​​​​​​​ \Rightarrow n - 5 = 0 $
$\Rightarrow n = 5$
Hence, $n = 5$

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