# Question #060f9

##### 2 Answers

#### Explanation:

Remember that:

#(""_b^a) = (a!)/(b!(a-b)!)#

Therefore, we can change our equation to say:

#(x!)/(2!(x-2)!) + ((x+1)!)/(2!(x-1)!) = x+6#

Simplifying both fractions, we get:

#((x)(x-1))/2 + ((x+1)(x))/2 = x+6#

#(x^2-x)/2 + (x^2 + x)/2 = x+6#

#(x^2-x+x^2+x)/2 = x+6#

#(2x^2)/2 = x+6#

#x^2 = x+6#

#x^2 - x - 6 = 0#

#(x-3)(x+2) = 0#

Therefore,

Since

The only answer left is

#x=3#

*Final Answer*

The answer is

#### Explanation:

The notation used is that of a combination of 2 objects chosen from a set of

I will first simplify the left side of the equation, then will combine this with the right side in order to solve.

The first combination is

or simply

The second combination is

or simply

Now, add the two expressions

So, the left side of the equation simplifies to

Now, the equation becomes

This factors as

The roots of this equation are

Since a combination having

The answer is