A group of 10 representatives is to be selected from 12 seniors and 10 juniors.\nIn how many different ways can the group be formed if it must contain at least one senior?

Introduction / Context:
This is a combination problem with a restriction. We must form a committee subject to a condition that at least one member comes from a particular subgroup. Questions of this type are very common in aptitude tests and are a direct application of the concept of combinations and the idea of counting by complement (total minus unwanted cases).

Given Data / Assumptions:

• Total seniors available = 12.
• Total juniors available = 10.
• Total people available = 22.
• We must select exactly 10 representatives.
• The selection must contain at least one senior.
• Order of members in the group does not matter, so combinations are used.

Concept / Approach:
The direct way would be to count all valid combinations that use at least one senior, but that requires splitting into many cases. A cleaner method is to first count all possible 10 person groups without any restriction and then subtract the single unwanted case where the group contains no senior at all. That unwanted group is the one which consists entirely of juniors. This is a classic example of using the complement method: valid groups = total groups - invalid groups.

Step-by-Step Solution:
Total people available = 12 seniors + 10 juniors = 22 people. Total ways to choose any 10 representatives from 22 people = C(22,10). Now consider invalid groups with no senior: then all 10 must be juniors. The number of ways to choose 10 juniors from the 10 juniors available = C(10,10) = 1. Therefore, valid groups with at least one senior = C(22,10) - C(10,10). Compute C(22,10) = 646646. So valid groups = 646646 - 1 = 646645.

Verification / Alternative check:
Notice that there is exactly one way to have zero seniors: taking all 10 juniors together. There is no other composition that avoids seniors while still having 10 members, because any group of size 10 that includes at least one senior would not be counted as invalid. This confirms that we really subtract only 1 case from the total C(22,10). The final result is therefore 646645, which is just one less than the unrestricted total.

Why Other Options Are Wrong:

• 646646: This is C(22,10) and ignores the restriction about including at least one senior.
• 497420: This is another combination value, such as C(22,9), which corresponds to choosing 9 rather than 10 members.
• 184756: This equals C(20,10) and comes from mistakenly assuming only 20 people instead of 22.

Common Pitfalls:
Students often try to enumerate all possible cases with different numbers of seniors, which is time consuming and error prone. Another frequent error is miscounting the invalid cases or assuming more than one way to have zero seniors. Some learners also confuse combinations and permutations and incorrectly multiply by factorials, which would give huge and meaningless numbers. Remember that for committees where order does not matter, combinations C(n,r) are the correct tool. Using the complement method is usually the simplest and safest approach when you see phrases like "at least one".

Final Answer:
The number of ways to form a group of 10 representatives containing at least one senior is 646645.