Problem 28 - A Complete Resource Book in Mathematics for JEE Main

Question

The numbers \(1,2,3, \ldots, n\) are arranged in a random order. The probability that the digits \(1,2,3, \ldots k\) ( \(k

Step-by-Step Solution

Answer

The probability is \(\frac{(n-k+1)!}{n!}\), which is option (D).

Step 1: Understand the Problem

We need to determine the probability that the numbers \(1, 2, 3, \ldots, k\) appear in order as sequential neighbors in a random permutation of \(1, 2, 3, \ldots, n\).

Step 2: Consider Placement as a Block

Think of the numbers \(1, 2, 3, \ldots, k\) as a single block rather than individual numbers. This block has \(k!\) different internal arrangements, but the sequence requires that they appear in ascending order.

Step 3: Count Total Arrangements

The total number of permutations of \(n\) numbers is \(n!\).

Step 4: Count Valid Arrangements

After treating \(1, 2, 3, \ldots, k\) as a single block, the sequence will have \((n-k+1)\) blocks in total if they are to be sequential. Thus, there are \((n-k+1)!\) ways to arrange these blocks.

Step 5: Calculate Probability

The probability is the number of valid arrangements divided by the total number of arrangements. Hence the probability is \(\frac{(n-k+1)!}{n!}\).

Step 6: Select Correct Answer

Based on the calculated probability, the correct answer is option (D), \(\frac{(n-k+1)!}{n!}\).

Key Concepts

Combinatorics

Combinatorics is a fundamental area of mathematics that deals with counting, arranging, and finding patterns within a set of elements. Whenever you encounter exercises involving permutations and combinations, you're working within the field of combinatorics. The key focus is on understanding how elements can be arranged in different ways, whether they are being selected or organized sequentially.
Combinatorics provides us with tools and formulas to solve complex problems that involve arrangements of objects. For instance:

• Permutations: Calculating the number of ways to arrange a set of objects. For a set of size \(n\), there are \(n!\) permutations.
• Combinations: Choosing a subset of objects, regardless of order. For choosing \(r\) objects from a set of \(n\), the formula is \(\binom{n}{r}\).

In problems like our original exercise, permutations are particularly useful because we are interested in the order in which numbers appear. Thus, an understanding of permutations helps us quickly determine the number of possible arrangements.

Arrangement of Numbers

The arrangement of numbers is a specific application of permutations in combinatorics. When we talk about arranging numbers, particularly in a line or sequence, we're interested in finding out how many different ways we can order them. Our goal is often to count these arrangements, considering constraints like being in ascending order or forming a particular sequence.
In the exercise you are solving, you need to find how many ways the numbers \(1, 2, 3, \ldots, k\) can appear as consecutive neighbors in a longer list of numbers, \(1, 2, 3, \ldots, n\).

• Consider these numbers as forming a "block." Treat this block as one single entity, which greatly simplifies the problem.
• Inside this block, the numbers \(1, 2, 3, \ldots, k\) must maintain their specific order since they are pre-defined in the exercise.

By thinking of it this way, the problem boils down to arranging this block with the remaining elements in multiple configurations, but always preserving the order inside the block itself.

Sequential Probability

Sequential probability refers to calculating the likelihood of events occurring in a specific order. Within a set sequence or permutation, we might want particular elements or events to occur in a prescribed sequence.
In the given problem, we want the numbers \(1, 2, 3, \ldots, k\) to be adjacent and in their natural order within the larger set of numbers \(1, 2, 3, \ldots, n\). To find this probability, we use:

• The total number of possible arrangements of \(n\) numbers, which is \(n!\).
• The number of valid arrangements where the block \(1, 2, 3, \ldots, k\) remains intact, calculated as \((n-k+1)!\).

The probability is the ratio of the number of valid arrangements to the total arrangements: \(\frac{(n-k+1)!}{n!}\). By understanding and applying the concepts of sequential probability, we have a clear pathway to solving these types of problems.