Question

If \(n\) objects are arranged in a row, then the number of ways of selecting three of these objects so that no two of them are next to each other is : (a) \({ }^{n} C_{3}\) (b) \({ }^{n-2} C_{3}\) (c) \({ }^{n} C_{2}\) (d) none of these

Short answer

Answer: \({ }^{n-2} C_{3}\)

Step-by-step solution

Visualize the objects and their positions

Imagine the n objects are placed in a row, and we want to choose three objects such that no two objects are next to each other. We can start by selecting the first object, and then move to the second and third objects.

Evaluate the available options

To make sure that none of the three objects are next to each other, we need to place a gap (at least one) between each pair of the selected objects. In other words, we need at least two gaps to separate the three objects. Thus, if we have n objects and three selections, we are left with n - 3 objects after placing the two gaps.

Count the number of possible selections

Now, we need to count the number of ways to select the three objects from the n - 3 objects left after placing the two gaps. This can be done using the combination formula, which is given by: C(n, r) = \(\frac{n!}{r!(n-r)!}\), where "n" is the total number of objects and "r" is the number of objects we want to choose. In our case, we want to choose 3 objects (r=3) from the n-3 available objects. So, our combination formula becomes: C(n-3, 3) = \(\frac{(n-3)!}{3!(n-3-3)!}\)

Simplify the combination formula

We can simplify the combination formula as follows: C(n-3, 3) = \(\frac{(n-3)!}{3!(n-6)!}\) This is equivalent to the option (b) in the given exercise: \({}^{n-2}C_{3}\). So, the number of ways of selecting three objects from a row of n objects such that no two of them are next to each other is given by the expression \({ }^{n-2} C_{3}\). The correct answer is thus (b).

Key concepts

Combination Formula

Understanding the combination formula is key to solving combinatorics problems. When faced with a question of selecting items from a set without caring about the order, this is where the combination formula comes into play.

It is written mathematically as C(n, r) = \(\frac{n!}{r!(n-r)!}\), where \(n\) is the number of items to choose from, \(r\) is the number of items to select, \(n!\) denotes the factorial of \(n\), and the exclamation mark '!' represents the factorial operation which multiplies the number by all the positive integers below it. For example, \(4! = 4 \times 3 \times 2 \times 1\).

Factorials grow rapidly, making the combination formula powerful for computing large possibilities in a concise expression. This is especially useful in the textbook exercise where we need to ensure no selected objects are adjacent, requiring us to tweak the usual formula.

Permutation and Combination

Permutation and combination are two fundamental concepts of combinatorics. Permutations focus on arranging objects where the order matters, while combinations deal with selecting objects with no regards for the sequence.

For permutations, expressed as \(P(n, r)\), every nuance in arrangement leads to a unique set. If you're creating a password using different characters, the order in which they're arranged is key. In combinations, however, where we use the aforementioned combination formula, the sequence of selection is irrelevant—what matters is simply which items are chosen.

The distinction is crucial when we consider the textbook problem at hand. We're looking at combinations because the solution does not require us to arrange the selected objects in a particular order, thereby reducing the number of possible configurations we have to count.

Arranging Objects

Arranging objects in different ways is a classic exercise in combinatorics, which requires careful consideration of restrictions, like ensuring certain objects aren't adjacent to each other.

In the given exercise, we're not merely selecting any three objects; we have to maintain gaps between them. This introduces an additional layer of complexity because we have to mentally insert these gaps into our row of objects before applying the combination formula. With these restrictions, the spaces between items become as crucial to our calculations as the items themselves.

By visualizing and structuring the problem, then employing the combination formula with our specific conditions, we can accurately deduce the number of valid selections. This approach to arranging objects with constraints is a technique that's widely applicable in more complex combinatorial problems.

Quantitative Aptitude

Quantitative aptitude involves the ability to handle numerical and logical reasoning. It is a crucial skill for solving problems in disciplines like mathematics, statistics, and computer science, and it's especially relevant in combinatorics.

The textbook exercise requires quantitative aptitude not just in applying mathematical formulas, but also in understanding the logical reasoning behind the arrangement of objects and the constraints involved. Students must read the problem, translate it into a mathematical model, and then manipulate this model according to the rules of combinatorics.

Enhancing quantitative aptitude can be achieved through regular practice and the careful breakdown of complex problems into simpler, manageable parts. By developing this skill, students will find it easier to approach a wide range of mathematical challenges with confidence.