While writing all the numbers from 700 to 1000, how many numbers occur in which the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place?

The correct option is 85.

Let the three-digit numbers be represented as $HTU$, where:
- $H$ represents the digit at the hundred's place.
- $T$ represents the digit at the ten's place.
- $U$ represents the digit at the unit's place.

We are looking for numbers in the range from 700 to 1000 (inclusive) that satisfy the condition:
$H>T>U$

Since we are looking at numbers from 700 to 1000:
- The number 1000 has $H=1$ (in terms of a 4-digit number representation, its thousand's digit is 1 and others are 0, which does not satisfy the three-digit decreasing condition). Thus, we only need to analyze three-digit numbers from 700 to 999.
- The hundred's digit $H$ can only be 7, 8, or 9.

Let's count the favorable numbers step-by-step for each possible value of $H$:

Case 1: When $H=7$
We need to find the number of pairs $(T,U)$ such that $7>T>U\ge 0$.
The possible digits for $T$ and $U$ must be chosen from the set of 7 digits: $\{0,1,2,3,4,5,6\}$.
Since $T>U$, any selection of 2 distinct digits from this set of 7 digits will automatically form a unique valid pair where the larger digit is assigned to $T$ and the smaller to $U$.
The number of ways to choose 2 digits from 7 is given by the combination formula $C^7$₂:

$C^7$₂=7×62=21

Case 2: When $H=8$
We need to find the number of pairs $(T,U)$ such that $8>T>U\ge 0$.
The digits for $T$ and $U$ must be chosen from the set of 8 digits: $\{0,1,2,3,4,5,6,7\}$.
The number of ways to choose 2 digits from 8 is:

$C^8$₂=8×72=28

Case 3: When $H=9$
We need to find the number of pairs $(T,U)$ such that $9>T>U\ge 0$.
The digits for $T$ and $U$ must be chosen from the set of 9 digits: $\{0,1,2,3,4,5,6,7,8\}$.
The number of ways to choose 2 digits from 9 is:

$C^9$₂=9×82=36

Total Count:
To find the total number of such numbers, we sum the counts from all three cases:

$\text{Total numbers}=21+28+36=85$

Thus, there are exactly 85 numbers between 700 and 1000 where the hundred's digit is greater than the ten's digit, and the ten's digit is greater than the unit's digit.