MadeEasy Workbook[2016] Q48

Question

Find the integer value of x satisfying the inequality$\binom{10}{x-1} < 2\binom{10}{x}$ .

Answer 1

$\frac{10!}{(x-1)!(10-(x-1))!}$  < 2$\frac{10!}{(x)!(10-x)!}$

$\frac{1}{(x-1)!(11-x)!}$           < $\frac{2}{(x)!(10-x)!}$

$\frac{1}{(11-x)}$                     < $\frac{2}{x}$

 x < 2(11-x)  =>   x < 22-2x  =>   3x < 22   =>   x < $\frac{22}{3}$

Comments

it is asking for an integr value and x=1 also satisfies the equation.

Answer 2

$\frac{10!}{(10-x+1)!*(x-1)!}<2\frac{10!}{(10-x)!*(x)!}$

$\frac{10!}{(10-x+1)*(10-x)!*(x-1)!}<2\frac{10!}{(10-x)!*(x)*(x-1))!}$

cancelling out common terms we get

$\frac{1}{(10-x+1)}$ < $\frac{2}{x}$

Equating it we get

x<$\frac{22}{3}$

so x>1 and x< 7

x={1,2,3,4,5,6,7}