<s>[INST] Probability to find 'X' number of Balls. A bag contains 5 Red, 4 Blue and x green balls. Two balls are drawn at random from the bag. If the probability of both being green is 1/7, find x.
I am clear with the proceedings but stuck at solving equation of factorials, so please show detailed steps.
 [/INST] Total balls = x + 9
Green balls = x
Probability of 2 balls both green $$= \frac{\binom x2}{\binom{x + 9}{2}}$$
$$\frac 17= \frac{\frac{x!}{2! × (x - 2)!}}{\frac{(x+9)!}{2! × (x + 7)!}}$$
$\frac 17= \frac{x!}{2! × (x - 2)!} × \frac{2! × (x + 7)!}{(x+9)!}$
$\frac 17= \frac{x!}{(x - 2)!} × \frac{(x+7)!}{(x+9)!}$
$\frac 17= \frac{x \cdot (x-1) \cdot (x-2)!}{(x - 2)!} × \frac{(x + 7)!}{(x+9)(x+8)(x+7)!}$
$\frac 17=  \frac {x \cdot (x-1)}{(x+9) \cdot (x+8)}$
$x^2 + 17x + 72 = 7x^2 - 7x$
$6x^2 - 24x -72 = 0$
$x^2 - 4x -12 = 0$
$x^2 -6x +2x - 12 = 0$
$x(x-6) +2(x-6) = 0$
$(x + 2 )(x - 6) = 0$
Either x = -2 or x = 6.
But x can't be negative.
So Green balls x = 6
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