To solve this problem, we are dealing with a combinations scenario because the order in which the subjects are selected doesn't matter. The formula for combinations is given by:

$C(n, k) = \frac{n!}{k!(n - k)!}$

Where:

• $n$ is the total number of subjects (44 in this case),
• $k$ is the number of subjects to be selected (4 here),
• $!$ represents factorial, meaning the product of all positive integers up to that number.

Step 1: Apply the formula

Here, $n = 44$ and $k = 4$, so the number of combinations is:

$C(44, 4) = \frac{44!}{4!(44 - 4)!} = \frac{44!}{4! \cdot 40!}$

Step 2: Simplify the factorials

We can cancel out the $40!$ from both the numerator and denominator:

$C(44, 4) = \frac{44 \times 43 \times 42 \times 41}{4 \times 3 \times 2 \times 1}$

Step 3: Compute the value

First, calculate the numerator:

$44 \times 43 = 1892$ $1892 \times 42 = 79464$ $79464 \times 41 = 3257024$

Now, calculate the denominator:

$4 \times 3 \times 2 \times 1 = 24$

Finally, divide the numerator by the denominator:

$C(44, 4) = \frac{3257024}{24} = 135710.5$

Thus, the number of different ways to select 4 subjects from 44 is 135,710.

Would you like more details or clarification? Here are some related questions you might find useful:

1. How do combinations differ from permutations?
2. How can this concept be applied to real-world scenarios like team selection?
3. How do you calculate combinations when the total number of subjects is large, like 100 or 1000?
4. What happens when the order of selection matters? (Hint: permutations)
5. Can you generalize the formula for combinations to more complex situations, like multiple groups of selections?

Tip: When calculating large combinations or factorials, always try to simplify the expression first (cancel out terms) before performing the multiplication and division.