If 1 C n 4 = 1 C n 5 + 1 C n 6 , then n =

Question

If $\frac{1}{{}^{4}C_{n}} = \frac{1}{{}^{5}C_{n}} + \frac{1}{{}^{6}C_{n}}$ , then $n =$

Smart Achievers · Accepted Answer

(2)

$\frac{1}{{}^{4}C_{n}} = \frac{1}{{}^{5}C_{n}} + \frac{1}{{}^{6}C_{n}}$ ; $\frac{1}{{}^{4}C_{n}} = \frac{{}^{6}C_{n} + {}^{5}C_{n}}{{}^{5}C_{n} {}^{6}C_{n}}$

$\Rightarrow {}^{5}C_{n} {}^{6}C_{n} = {}^{4}C_{n} [{}^{6}C_{n} + {}^{5}C_{n}]$

$\Rightarrow \frac{5! 6!}{(5 - n)! (6 - n)! {(n!)}^{2}} = \frac{4!}{(4 - n)! n!} [{}^{6}C_{n} + {}^{5}C_{n}]$

$\Rightarrow \frac{5 \cdot 6}{(5 - n) (6 - n)} = \frac{6}{6 - n} + 1 = \frac{12 - n}{6 - n}$

$\Rightarrow (5 - n) (12 - n) = 30$ $\Rightarrow n^{2} - 17 n + 60 = 30$

$\Rightarrow n^{2} - 17 n + 30 = 0$

$∴$ $n = 15, 2$ $[15 is not possible (∵ n \leq 4)]$

$∴$ $n = 2$

Solution

Q.
If $\frac{1}{{}^{4}C_{n}} = \frac{1}{{}^{5}C_{n}} + \frac{1}{{}^{6}C_{n}}$ , then $n =$

1 3

2 2

3 1

4 0

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Solution

Q. If 1Cn4=1Cn5+1Cn6, then n=

1 3

2 2

3 1

4 0

Ans. (2) 1Cn4=1Cn5+1Cn6; 1Cn4=Cn6+Cn5Cn5 Cn6 ⇒Cn5 Cn6=Cn4[Cn6+Cn5] ⇒5! 6!(5-n)!(6-n)!(n!)2=4!(4-n)! n![Cn6+Cn5] ⇒5·6(5-n)(6-n)=66-n+1=12-n6-n ⇒(5-n)(12-n)=30⇒n2-17n+60=30 ⇒n2-17n+30=0 ∴ n=15, 2 [15 is not possible (∵ n≤4)] ∴ n=2

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Q.
If $\frac{1}{{}^{4}C_{n}} = \frac{1}{{}^{5}C_{n}} + \frac{1}{{}^{6}C_{n}}$ , then $n =$