2C_4 ^n = 35C_3 ^(n/2) ⇒ n = ?


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Question Number 74129 by malwaan last updated on 19/Nov/19

$2 C_{4}^{n} = 35 C_{3}^{\frac{n}{2}}$ $\Rightarrow n = ?$
	Answered by MJS last updated on 20/Nov/19

	$n \in N \Rightarrow n ⩾ 4 \land \frac{n}{2} ⩾ 3 \Rightarrow n ⩾ 6$ $n = 2 k (k ⩾ 3)$ $2 C_{4}^{2 k} = 35 C_{3}^{k}$ $2 \frac{(2 k)!}{4! (2 k - 4)!} = 35 \frac{k!}{3! (k - 3)!}$ $\frac{1}{3} k (k - 1) (2 k - 3) (2 k - 1) = \frac{35}{6} k (k - 2) (k - 1)$ $k = 0 \lor k = 1 not valid$ $\frac{1}{3} (2 k - 3) (2 k - 1) = \frac{35}{6} (k - 2)$ $2 (2 k - 3) (2 k - 1) = 35 (k - 2)$ $8 k^{2} - 51 k + 76 = 0$ $(k - 4) (8 k - 19) = 0$ $k = \frac{19}{8} not valid$ $k = 4$ $\Rightarrow n = 8$
	Commented by malwaan last updated on 20/Nov/19

	$t h a n k y o u s o m u c h s i r M J S$
	Commented by MJS last updated on 20/Nov/19

	${you}^{'} re welcome$
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