Question

if nC2 -nC1=35 THEN FIND THE VALUE OF N ?

Answer 1

hence n value is 10....

Answer 2

The value of n is 10.

Step-by-step explanation:

We have,

$^{n} C_2-^{n} C_1=35$

To find, the value of n = ?

∴$^{n} C_2-^{n} C_1=35$

⇒ $\dfrac{n!}{2!(n-2)!} -\dfrac{n!}{1!(n-1)!}=35$

[ ∵ $^{n} C_r=\dfrac{n!}{r! (n-r)!}$]

⇒ $\dfrac{(n-)n(n-2)!}{2!(n-2)!} -\dfrac{(n-1)!n}{1!(n-1)!}=35$

⇒$\dfrac{(n-)n}{2} -\dfrac{n}{1}=35$

⇒$n^{2} -n-2n=70$

⇒$n^{2} -3n-70=0$

⇒$n^{2} -10n+7n-70=0$

⇒ $(n-10)(n+7)=0$

⇒ n = 10 or - 7 [n never be negative]

∴ n = 10

Hence,  the value of n is 10.