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When working with integers, it's essential to understand the difference between even and odd numbers. Even integers are those that can be divided exactly by two. They have no remainder. For example, numbers like 2, 4, and 6 are all even because they satisfy this condition. Conversely, odd integers do not divide evenly by two, leaving a remainder of one. Examples include numbers like 1, 3, and 5.

In the context of our problem, we dealt with the set of numbers from 1 to 101. Within this set, identifying which numbers are even or odd was crucial. By observing the sequence, every second number starting from 2 is even, so we have even integers: 2, 4, 6, ..., 100. Similarly, starting with 1, every subsequent odd-positioned number like 1, 3, 5, ..., 101 is odd. Consequently, there are 50 even numbers and 51 odd numbers in the set.

In combinatorics, combinations without repetition are a fundamental concept. They help calculate the number of different groups that can be made from a larger set, where order does not matter, and each element can only be used once.

The problem requires us to find pairs of integers from a set, specifically those pairs whose sum is even. We achieve this by selecting either two even numbers or two odd numbers, as the sum of two even or two odd integers results in an even number.

• We use the formula for combinations, given by: \[ C(n, k) = \frac{n!}{k!(n - k)!} \]
• Here, \( n \) represents the total number of items, and \( k \) is the number of items to choose.
• The exclamation mark denotes "factorial," which we'll explain in the following section.

By applying this combination formula, we computed the number of ways two even numbers can pair up, and separately, two odd numbers, to maintain their even sum characteristic.

Factorials are a cornerstone in combinatorics, denoted by \(!\). The factorial of a number \( n \), represented as \( n! \), is the product of all positive integers up to \( n \). Thus, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

When calculating combinations, factorial simplification is key to simplifying expressions. This is crucial when dealing with large numbers, as direct computation can be cumbersome.

• In our solution, we used \( \frac{50!}{2!48!} \) and \( \frac{51!}{2!49!} \) to determine the number of ways pairs could form.
• By canceling out common factorial terms from the numerator and the denominator, we transformed these expressions into simpler multiplications: \( \frac{50 \times 49}{2} \) and \( \frac{51 \times 50}{2} \).
• This simplification reduced the complexity significantly, allowing us to swiftly find the number of desired pairs.

Ultimately, simplifying factorials helped compute and sum the number of pairs in a manageable manner, leading to the final answer.