From a set of 20 distinct objects, how many different groups of 5 objects can be formed if the order of objects inside each group does not matter at all?

Introduction / Context:
This question tests your understanding of basic combinatorics, specifically the concept of combinations. When we form groups from a larger set and do not care about the internal order of items in each group, we use combinations, not permutations. Such questions are very common in aptitude exams, engineering entrance tests, and competitive exams, so it is important to clearly know when to use nCr instead of nPr.

Given Data / Assumptions:

• There are 20 distinct objects in total.
• We must choose groups containing exactly 5 objects.
• The order of objects within each group does not matter.
• Each object can either be in a group or not, and no object is repeated inside a group.

Concept / Approach:
To count selections where order does not matter, we use combinations. The formula for combinations is:
nCr = n! / (r! * (n - r)!) Here n = 20 and r = 5. We simply substitute in this formula and simplify. This gives the exact number of different 5-object groups that can be formed from 20 distinct objects.

Step-by-Step Solution:
Step 1: Identify n = 20 (total objects) and r = 5 (objects in one group). Step 2: Use the formula nCr = n! / (r! * (n - r)!). Step 3: Compute 20C5 = 20! / (5! * 15!). Step 4: Expand only what is needed: 20C5 = (20 * 19 * 18 * 17 * 16) / (5 * 4 * 3 * 2 * 1). Step 5: Simplify the numerator and denominator carefully: (20 * 19 * 18 * 17 * 16) = 1860480 and 5! = 120. Step 6: Divide: 1860480 / 120 = 15504. Step 7: Therefore, the number of distinct 5-object groups is 15504.

Verification / Alternative check:
We can quickly check reasonableness. 20C1 is 20, 20C2 is 190, and combinations grow larger as r increases up to around the middle. Since 20C5 is symmetric with 20C15 and is near the central region, a value in the tens of thousands is reasonable. Also, a calculator or combinatorics table confirms that 20C5 = 15504, so our result is consistent with standard values.

Why Other Options Are Wrong:
57090: This does not match the correct application of the combination formula and is too large. 15540: Very close in magnitude but results from incorrect simplification or arithmetic. 23670: Again, this does not correspond to 20C5 and comes from an incorrect calculation. These distractors are typical of small arithmetic mistakes in factorial simplification.

Common Pitfalls:
Many students confuse permutations and combinations. Using nPr instead of nCr would wrongly treat different orders of the same group as different arrangements. Another frequent mistake is partial cancellation of factorial terms, leading to minor numerical errors like 15540 instead of 15504. It is important to carefully cancel common terms in numerator and denominator and to compute step by step for accuracy. Also, forgetting that order does not matter here leads to overcounting and a much higher incorrect number.

Final Answer:
The total number of distinct groups of 5 objects that can be formed from 20 different objects, when order inside each group does not matter, is 15504.