In basic combinatorics, what is the value of the binomial coefficient 6 choose 3, which counts the number of ways to choose 3 objects from 6 distinct objects?

Introduction / Context:
The expression 6 choose 3, written as 6C3 or C(6, 3), appears frequently in combinatorics, probability and many aptitude tests. It represents the number of distinct subsets of size 3 that can be formed from a set of 6 distinct elements, when order does not matter. Understanding how to evaluate such expressions using factorials is a fundamental skill.

Given Data / Assumptions:
- We have 6 distinct objects in total.
- We want to select 3 of these objects at a time.
- The order of selection does not matter, only the chosen group matters.
- We are asked to compute the value of 6C3.

Concept / Approach:
The binomial coefficient nC r is defined by the formula nC r = n! / (r! * (n - r)!). For 6C3, n is 6 and r is 3. Using factorial values 6! = 720 and 3! = 6, and (6 - 3)! = 3! = 6, we can substitute into the formula and simplify. Often, instead of computing full factorials, we cancel common factors to make the calculation quicker and less error prone.

Step-by-Step Solution:
Step 1: Write the general formula for combinations: nC r = n! / (r! * (n - r)!). Step 2: Substitute n = 6 and r = 3 to get 6C3 = 6! / (3! * 3!). Step 3: Expand 6! as 6 * 5 * 4 * 3 * 2 * 1 and 3! as 3 * 2 * 1. Step 4: Cancel common factors to simplify: 6! / (3! * 3!) = (6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (3 * 2 * 1)). Step 5: After cancellation, you can simplify directly: 6C3 = (6 * 5 * 4) / (3 * 2 * 1) = 120 / 6 = 20. Step 6: Conclude that there are 20 distinct ways to choose 3 objects out of 6.

Verification / Alternative Check:
As a check, you might list combinations in a systematic way, such as fixing the smallest element and choosing pairs for the rest. Although listing becomes impractical for large n, for small n like 6, you can count and confirm that exactly 20 unique triplets are possible. This matches the value found using the formula and confirms the calculation.

Why Other Options Are Wrong:
Option B (30) could come from confusing combinations with permutations or from miscalculations in factorial simplification.
Option C (18) may result from incorrect cancellation or forgetting one combination.
Option D (24) is typical when someone mistakenly uses permutations 4P3 or another unrelated expression instead of 6C3.

Common Pitfalls:
Learners sometimes confuse nC r with nP r, or they fail to cancel factorial terms correctly. Another common issue is thinking that order matters for combinations, which is not correct. Remember that combinations count selections where the arrangement does not matter.

Final Answer:
The value of the binomial coefficient 6 choose 3 is 20.