Problem 26 - Algebra for JEE ADVANCED

Question

The number of different seven digit numbers that can be written using only the three digits 1,2 and 3 with the condition that the digit 2 occurs twice in each number is a. \({ }^{7} P_{2} 2^{5}\) b. \(\quad{ }^{7} C_{2} 2^{5}\) c. \({ }^{7} C_{2} 5^{2}\) d. none of these

Step-by-Step Solution

Answer

The correct answer is b. \( ^{7}C_{2} \times 2^5 \).

Step 1: Understand the Problem Requirements

We're asked to find the number of different seven-digit numbers using only the digits 1, 2, and 3, with the condition that exactly two of the digits are '2'. This limits the other five digits to 1 or 3.

Step 2: Identify the Number of '2's

The number '2' must appear exactly twice in the seven-digit number. We have to select 2 out of the 7 positions for the digit '2' to occur.

Step 3: Calculate the Number of Ways to Position '2's

To calculate the number of ways to choose 2 positions from 7 for the digit '2', we use combinations: \[ \binom{7}{2} \]

Step 4: Determine the Remaining Positions

After placing the two '2's, there are 5 positions left that can be filled with either digit '1' or '3'.

Step 5: Calculate Total Combinations for Remaining Positions

Each of these 5 positions can be either '1' or '3'. Since each position has 2 choices, the total number of combinations for these positions is \[ 2^5 \]

Step 6: Calculate Total Number of Combinations

Multiply the number of ways to choose positions for '2's by the number of combinations possible for the remaining digits:\[ \binom{7}{2} \times 2^5 \]

Step 7: Answer the Exercise

\( \binom{7}{2} = 21 \) and \( 2^5 = 32 \). Multiply these two results:\[ 21 \times 32 = 672 \]Thus, the answer is option 'b', \( ^{7}C_{2} \times 2^5 \).

Key Concepts

Permutations and Combinations

Permutations and combinations are essential concepts in combinatorics, a field of mathematics focused on counting and arrangement principles. When dealing with permutations, we are generally concerned about the arrangement of objects in a specific order. In contrast, combinations deal with selecting objects without regard to order.

In our exercise, we used combinations to determine how to position the specific digit '2' in the seven-digit number. Here, the order of placing digits is not important—only the positions matter. Therefore, we apply the combination formula \[\binom{n}{r} = \frac{n!}{r!(n-r)!}\] where \(n\) is the total number of positions, and \(r\) is the number we choose. In this case, it's about choosing 2 positions from 7 so \(\binom{7}{2}\).

• Permutation: Arrangement is crucial.
• Combination: Order is not important.

Understanding combinations and their difference from permutations helps us solve various counting problems with clarity.

Binomial Theorem

The binomial theorem is a foundational principle in algebra, describing the expansion of powers of a binomial. This theorem simplifies the process of expanding expressions like \((x + y)^n\), using coefficients represented by binomial coefficients \(\binom{n}{k}\).

In the context of our problem, while the binomial theorem is not explicitly used, its concept of combinations and powers is essential. We use combinations to select positions for '2's, and each remaining position for the digits '1' or '3' follows a simple power rule: \(2^5\). This power rule is analogous to the terms \((x+y)^5\) and represents the various ways 2 digits can fill the remaining positions.

The theorem shows how these expressions can be simplified and calculated efficiently. It brings an understanding of how combinations relate to algebraic expressions and provides a structure to calculate potential outcomes.

Probability

Probability measures the likelihood of a particular outcome occurring among all possible results. Although our exercise primarily involves combinatorics for counting arrangements, understanding probability connects directly to these counting principles.

By determining all possible outcomes using combinations in our example, we can, in theory, calculate probabilities. For instance, if someone asked, "What is the probability of forming such a seven-digit number under these conditions?" it would involve listing all valid combinations and finding how many of them fit the criteria. The probability formula\( P(A) = \frac{|A|}{|S|} \) could be applied once we know all the successful and possible outcomes.

• Counting outcomes: We calculated a total of 672 valid numbers based on our conditions.
• Probability connection: It provides a basis if probability questions arise.

So, probability concepts ensure we understand the likelihoods based on these calculated outcomes, even if not directly required in solving the exercise itself.