For a 4-digit number (greater than 1000), sum of the digits in the thousands, hundreds, and tens places is 15. Sum of the digits in the hundreds, tens, and units places is 16. Also, the digit in the tens place is 6 more than the digit in the units place. The difference between the largest and smallest possible value of the number is

The correct option is 811.

Let the 4-digit number be represented as $abcd$, where:
$a$ is the digit in the thousands place ($1\le a\le 9$),
$b$ is the digit in the hundreds place ($0\le b\le 9$),
$c$ is the digit in the tens place ($0\le c\le 9$),
$d$ is the digit in the units place ($0\le d\le 9$).

From the given problem, we can write the following equations based on the conditions:

Condition 1: Sum of the digits in the thousands, hundreds, and tens places is 15.

$a+b+c=15$

Condition 2: Sum of the digits in the hundreds, tens, and units places is 16.

$b+c+d=16$

Condition 3: The digit in the tens place is 6 more than the digit in the units place.

$c=d+6$

Now, let us determine the possible values for the digits $d$, $c$, $b$, and $a$:

Since $c$ is a single-digit number ($c\le 9$), and $c=d+6$, the possible values for $d$ can be 0, 1, 2, or 3.

Subtracting Condition 1 from Condition 2 gives:

$(b+c+d)-(a+b+c)=16-15$

$d-a=1$
$a=d-1$

Since $a$ must be a non-zero digit ($a\ge 1$), $d$ must be at least 2 ($d\ge 2$).

Thus, the possible values for $d$ are 2 and 3.

Case 1: When $d=2$

$a=2-1=1$
$c=2+6=8$
$b=15-a-c=15-1-8=6$

So, the 4-digit number is 1682.

Case 2: When $d=3$

$a=3-1=2$
$c=3+6=9$
$b=15-a-c=15-2-9=4$

So, the 4-digit number is 2493.

Now, we find the largest and smallest possible values among the numbers formed:

Largest possible value = 2493
Smallest possible value = 1682

The difference between the largest and smallest possible value is:

$2493-1682=811$