# If... color(white)x^(x + 1)C_3 - ^(x - 1)C_3 = 16x = ?

Nov 22, 2017

$x = 5$

#### Explanation:

${C}_{3}^{x + 1} = \frac{\left(x + 1\right) x \left(x - 1\right)}{1.2 .3}$

and ${C}_{3}^{x - 1} = \frac{\left(x - 1\right) \left(x - 2\right) \left(x - 3\right)}{1.2 .3}$

Therefore ${C}_{3}^{x + 1} - {C}_{3}^{x - 1}$

= $\frac{\left(x + 1\right) x \left(x - 1\right)}{1.2 .3} - \frac{\left(x - 1\right) \left(x - 2\right) \left(x - 3\right)}{1.2 .3}$

= $\frac{x - 1}{6} \left\{x \left(x + 1\right) - \left(x - 2\right) \left(x - 3\right)\right\}$

= $\frac{x - 1}{6} \left\{{x}^{2} + x - \left({x}^{2} - 5 x + 6\right)\right\}$

= $\frac{x - 1}{6} \left(6 x - 6\right)$

= ${\left(x - 1\right)}^{2}$

Hence ${C}_{3}^{x + 1} - {C}_{3}^{x - 1} = 16 x$ is equivalent to

${\left(x - 1\right)}^{2} = 16$

or $x - 1 = \pm 4$

i.e. $x = 5$ or $- 3$

and if we consider only as a positive integer $x = 5$