Problem 12 - Algebra for JEE ADVANCED

Question

Total number of six-digit numbers that can be formed, having the property that every succeeding digit is greater than the preceding digit, is equal to a. \({ }^{9} \mathrm{C}_{3}\) b. \({ }^{10} C_{3}\) c. \({ }^{9} p_{3}\) d. \({ }^{10} p_{3}\)

Step-by-Step Solution

Answer

The solution is option a: \(^{9}C_{3} = 84\).

Step 1: Understanding the Problem

We need to find the total number of six-digit numbers where each digit is greater than the one before it. This implies that each choice introduces a new digit without repetition. Thus, the problem requires finding the combination of selecting 6 distinct digits in increasing order from the set {1, 2, ..., 9}.

Step 2: Counting Combinations

Since any number formed must be strictly increasing, it involves choosing any 6 distinct digits out of the possible set {1, 2, ..., 9}. The condition implies choosing combinations without concern for order within the chosen set.

Step 3: Applying Combinatorial Formula

We employ the formula for combinations, which is \ {n \choose k} = \frac{n!}{k!(n-k)!}. Here, we need to calculate \(^{9}C_{6}\). This is equal to \(^{9}C_{3}\) because \(^{n}C_{k} = ^{n}C_{n-k}\).

Step 4: Final Calculation and Answer

Compute using the formula: \(^{9}C_{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\). Thus, option a, which is \(^{9}C_{3}\), is the correct answer.

Key Concepts

Combinations

In combinatorics, combinations refer to the selection of items from a larger pool, where the order does not matter. Think of it like picking a hand of cards from a deck; the sequence in which you pull them out does not change your hand's composition.

When solving a problem that involves combinations, the idea is to determine how many ways we can choose a subset of items from a set without regard to the order. The mathematical formula for combinations is:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

where \(n\) is the total number of items, \(k\) is the number of items to choose, and \(!\) denotes factorial, the product of all positive integers up to that number.

In the context of the exercise, selecting 6 digits such that each subsequent digit is greater than the previous dictates an inherent order of increasing sequences. Therefore, the task simplifies to choosing any 6 distinct digits from a potential pool of 9, leading us to apply combinations.

Permutations

Permutations differ from combinations as they consider the order of selection. This is akin to arranging books on a shelf, where the order in which they are placed makes a difference.

The formula for permutations when choosing \(k\) items from \(n\) is:

• \( P(n, k) = \frac{n!}{(n-k)!} \)

Notice that permutations do not divide by \(k!\) as combinations do; this is because the sequence does matter in permutations.

In the problem presented in the exercise, the digits need to be increasing, and thus, permutations are not applicable here because combinations already address letting the order be increasing automatically.

Increasing sequences

Increasing sequences are lists of numbers where each element is larger than the one that came before. This is a straightforward yet specific requirement that results in a unique ordering, naturally aligning with the essence of combinations rather than permutations.

For the exercise, because the six-digit number needs each digit to exceed the last, you're essentially forming an increasing sequence. Choosing 6 out of 9 digits and arranging them in increasing order automatically adheres to such a sequence.

Thus, the challenge becomes a matter of how many such sequences are possible, translating into a problem of combinations. While the order of the numbers is crucial in the formation of an increasing sequence, it is not an additional choice but a built-in condition of choosing these digits.