We are given the expression $\frac{3 \cdot X! + 15\cdot 4!}{6!}$ and we know the value of the expression is 1.
To find the value of $X$, we can simplify the expression and solve for $X$.
First, let's expand the factorials in the expression:
$\frac{3 \cdot X \cdot (X-1) \cdot (X-2) \cdot ... \cdot 3 \cdot 2 \cdot 1 + 15 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$
Simplifying the numerator, we have:
$\frac{3 \cdot X! + 15 \cdot 4!}{6!} = \frac{3 \cdot X \cdot (X-1) \cdot (X-2) \cdot ... \cdot 3 \cdot 2 \cdot 1 + 15 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$
$\frac{3 \cdot X! + 15 \cdot 4!}{6!} = \frac{3 \cdot X \cdot (X-1) \cdot (X-2) \cdot ... \cdot 3 \cdot 2 + 15 \cdot 4 \cdot 3 \cdot 2}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$
Now, let's simplify the expression further:
$\frac{3 \cdot X! + 15 \cdot 4!}{6!} = \frac{3 \cdot X \cdot (X-1) \cdot (X-2) \cdot ... \cdot 3 \cdot 2 + 15 \cdot 4 \cdot 3 \cdot 2}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$
$\frac{3 \cdot X! + 15 \cdot 4!}{6!} = \frac{3 \cdot X \cdot (X-1) \cdot (X-2) \cdot ... \cdot 3 \cdot 2 + 15 \cdot 4 \cdot 3 \cdot 2}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$
$\frac{3 \cdot X! + 15 \cdot 4!}{6!} = \frac{3 \cdot X \cdot (X-1) \cdot (X-2) \cdot ... \cdot 3 \cdot 2 + 15 \cdot 4 \cdot 3 \cdot 2}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$
Now, we know that the value of the expression is 1, so we can set up the equation:
$\frac{3 \cdot X \cdot (X-1) \cdot (X-2) \cdot ... \cdot 3 \cdot 2 + 15 \cdot 4 \cdot 3 \cdot 2}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 1$
To solve for $X$, we can simplify the equation:
$\frac{3 \cdot X! + 15\cdot 4!}{6!} = 1$
$\frac{3 \cdot X! + 15\cdot 24}{720} = 1$
$\frac{3 \cdot X! + 360}{720} = 1$
$3 \cdot X! + 360 = 720$
$3 \cdot X! = 720 - 360$
$3 \cdot X! = 360$
Dividing both sides of the equation by 3, we find:
$X! = 120$
The value of $X$ is 5.
The answer is: 5
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Calculate $\frac{3 \cdot X! + 15\cdot 4!}{6!}$
If we know the answer to the above question is 1, what is the value of unknown variable X?

### Answer:
