How many numbers are there between 100 and 300 which either begin with or end with 2 ?

The correct option is 110.

To find the number of integers between 100 and 300 (exclusive of 100 and 300, i.e., from 101 to 299) that either begin with 2 or end with 2, we can apply the Principle of Inclusion-Exclusion.

Let $A$ be the set of numbers that begin with 2.
Let $B$ be the set of numbers that end with 2.

We want to find the number of elements in the union of these two sets, which is given by:
$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$

Step 1: Find the numbers that begin with 2 (Set A)
The numbers between 100 and 300 that begin with 2 are the three-digit numbers where the hundreds digit is 2. These numbers range from 200 to 299 inclusive.
The number of such integers is:
$n\left(A\right)=299-200+1=100$

Step 2: Find the numbers that end with 2 (Set B)
The numbers between 100 and 300 (from 101 to 299) that end with 2 must have a units digit of 2. These are of the form _ _ 2.
Since the numbers are between 100 and 300, the hundreds digit can be 1 or 2.
- Case 2.1: If the hundreds digit is 1 (numbers from 101 to 199), the number is of the form 1 _ 2. The tens digit can be any digit from 0 to 9 (10 options). This gives 10 numbers.
- Case 2.2: If the hundreds digit is 2 (numbers from 200 to 299), the number is of the form 2 _ 2. The tens digit can be any digit from 0 to 9 (10 options). This gives 10 numbers.
Total numbers ending with 2:
$n\left(B\right)=10+10=20$

Step 3: Find the numbers that both begin with 2 and end with 2 (Set A ∩ B)
These are the numbers of the form 2 _ 2. The hundreds digit is 2, and the units digit is 2. The tens digit can be any digit from 0 to 9.
Therefore, the number of such integers is:
$n(A\cap B)=10$

Step 4: Calculate the final union
Using the formula:
$n(A\cup B)=100+20-10=110$

Thus, there are exactly 110 numbers between 100 and 300 which either begin with or end with 2.