Problem 44 - Algebra for JEE ADVANCED

Question

In how many ways can a team of six horses be selected out of a stud of 16 , so that there shall always be three out of \(A B C A^{\prime} B^{\prime}\) \(C^{\prime}\), but never \(A A^{\prime}, B B^{\prime}\) or \(C C^{\prime}\) together a. 840 b. 1260 c. 960 d. 720

Step-by-Step Solution

Answer

The answer is 960.

Step 1: Understand the Problem

We need to select a team of 6 horses from a group of 16. Specifically, this team must include 3 horses from the subset \( \{ A, B, C, A', B', C' \} \), ensuring that the paired horses \( A \) and \( A' \), \( B \) and \( B' \), \( C \) and \( C' \) are not selected together.

Step 2: Determine Combinations for the Pairs

For the group \( \{ A, B, C, A', B', C' \} \), we can't have both \( A \) and \( A' \), \( B \) and \( B' \), or \( C \) and \( C' \) together. There are 2 choices for each pair: either the original or the alternate. Therefore, there are \( 2^3 = 8 \) such combinations.

Step 3: Calculate Combinations with the Remaining Horses

After choosing 3 out of \( \{ A, B, C, A', B', C' \} \), we need to select 3 more horses from the remaining 10 horses (as there are 16 total and 6 have been partially fixed). This can be calculated using the combination formula \( \binom{n}{k} \). The combinations are \( \binom{10}{3} = 120 \).

Step 4: Combine the Results

Each valid selection from step 2 can be combined with any of the selections from step 3. Therefore, the total number of valid selections is \( 8 \times 120 = 960 \).

Step 5: Conclusion

The total number of ways to select the horses ensuring the conditions are met is 960.

Key Concepts

Combination formula

In combinatorics, the combination formula is a fundamental tool that helps us determine how many ways we can select a subset of items from a larger set, without taking the order of selection into account. This is often denoted as \( \binom{n}{k} \), where \( n \) is the total number of items to pick from, and \( k \) is the number of items to actually select. The formula is expressed as:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]This formula calculates the number of possible combinations by dividing the total number of permutations of \( n \) items by the number of ways to arrange the \( k \) selected items and the \( n-k \) unselected items. In our exercise, after securing three compliant horses from the subset \( \{A, B, C, A', B', C'\} \), we then calculated the number of ways to choose the remaining horses from the other 10 using this combination formula: \( \binom{10}{3} = 120 \). Understanding how to apply the combination formula is crucial in various combinatorial problems.

Team selection

Team selection problems often involve choosing a subset from a larger group. However, conditions or restrictions can add layers of complexity, as seen in our exercise with horse selection.

• Firstly, we have to ensure that three horses are selected from a specific subgroup forming part of the larger team, specifically \( \{ A, B, C, A', B', C' \} \).
• Secondly, there are additional constraints - in this case, ensuring no pair \( A \) and \( A' \), \( B \) and \( B' \), or \( C \) and \( C' \) can appear together.

Because of these restrictions, a standard approach is first to evaluate how to choose those complying with constraints, and subsequently fill the remaining spots freely from the unbounded portion of the group. This often involves initially fixing some parts of the team based on the problem's conditions and using combinations for the remainder as we did with the 10 horses. By tackling selections methodically, one can ensure all conditions are met satisfactorily.

Permutations and combinations

Permutations and combinations are two closely related concepts in combinatorics, both useful for counting arrangements of elements.

• A permutation considers order—how many ways you can arrange a certain group of items.
• A combination, on the other hand, ignores order—focusing only on selecting items from a group regardless of arrangement.

In the context of our exercise, combinations are appropriate because the order in which we select horses doesn't matter. Our task is simply to determine how many distinct sets of horses can be chosen under given constraints. Permutations would have been used if we needed to calculate specific sequences, such as arranging the selected horses in a racing order. By knowing when to apply permutations versus combinations, one can tackle a wide array of mathematical problems involving group selections.

Disjoint sets

Disjoint sets in mathematics are sets that have no elements in common. This concept is often crucial for ensuring that selected groups meet specific criteria, such as avoiding unwanted overlaps or pairings. In our problem, although we did not have outright disjoint sets, we aimed to avoid certain pairings among the horses—namely, the simultaneous selection of any pair \( (A, A'), (B, B'), (C, C') \).

• We ensured compliance by imposing restrictions on combinations from \( \{ A, B, C, A', B', C' \} \).
• This constraint effectively operated as if these pairs were treated as disjoint—choosing one horse in a pair excluded the alternative.

Using these kinds of restrictions imitates managing disjoint sets, as it prevents overlap in selection. Understanding disjoint sets and how to apply analogous principles helps manage complex constraints while forming valid subgroups or solving selection problems effectively.