Problem 69 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The number of different 7 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) \({ }^{7} P_{2} 2^{2}\) (B) \({ }^{7} C_{2} 2^{5}\) (C) \({ }^{7} C_{2} 5^{5}\) (D) None of these

Step-by-Step Solution

Answer

672 numbers can be made.

Step 1: Understand the Problem

We are tasked with finding the number of 7-digit numbers using the digits 1, 2, and 3 with the condition that the digit 2 appears exactly twice. Therefore, we need to determine how many combinations of the digits 1, 2, and 3 can form such numbers.

Step 2: Choose Positions for Two '2's

Determine where the two '2's will be placed among the 7 digits. We use combinations to select 2 positions out of 7 for the '2's. The number of ways to do this is given by the combination formula: \[{}^{7}C_{2} = \frac{7!}{2!(7-2)!} = 21\]

Step 3: Fill Remaining Positions with '1' or '3'

After fixing the positions for '2's, there remain 5 positions that can be filled with either '1' or '3'. Each position has 2 choices, which gives us \[2^5 = 32\] different ways to fill these positions.

Step 4: Calculate Total Number of 7-digit Numbers

Multiply the number of ways to position the '2's by the number of ways to fill the remaining positions. Thus, the total number of 7-digit numbers is:\[21 \times 32 = 672\]Thus, OPTION (B) \({ }^{7} C_{2} 2^{5}\) = 672 is not matching the result directly, but values themselves are used in the needed formula.

Key Concepts

Combinations

In combinatorics, combinations are a way to select a group of items from a larger set without considering the order of the items. When we talk about combinations, the order in which we pick the items doesn't matter. The mathematical formula to calculate combinations is written as \( ^nC_r \), which represents the number of ways to choose \( r \) items from \( n \) total items, and it is calculated by:

• \( ^nC_r = \frac{n!}{r!(n-r)!} \)

Here, \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \).
In the context of the exercise, the combinations are used to select positions for the digit '2' in the number.
We have 7 positions and need to choose 2 of them for the digit '2', which is calculated as \( ^7C_2 = 21 \).
This step shows us how we focus merely on selecting positions without stressing about sequential order.

Permutations

Permutations differ from combinations as they focus on the order of selection. In permutations, the sequence matters, and that is critical when arrangements concern order-specific outcomes.
The formula for permutations is denoted as \( ^nP_r \), corresponding to the number of ways to arrange \( r \) items from \( n \) distinct items:

• \( ^nP_r = \frac{n!}{(n-r)!} \)

While permutations are essential in circumstances where the order cannot be ignored, our problem stressed the placement requirements using combinations, emphasizing positions only.
The exercises imply permutations for other possible arrangements, but they concentrate more on achieving the numbers regardless of arrangements. Although permutations didn't play a major role in the solution here, understanding both concepts allows for a fuller grasp of arrangements and selections.

Binomial Theorem

The binomial theorem provides a powerful formula to expand expressions raised to a power, expressed as \((a + b)^n\). It demonstrates how any power of a binomial can be expanded into a sum involving terms of the form \( nCk a^{n-k} b^k \).

• \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]

Each term in the expansion corresponds to a certain selection of \( a \)s and \( b \)s, akin to partitioning an array of items into orders of selection.
In the exercise, the underlying principles of combinations are reflective of this theorem, especially where they mold how we select positions for our digits '1', '2', and '3'.
By looking at combinations and the positional arrangements in this problem, the binomial theorem subtly nudges us to understand how larger expressions factor into component combinations, similar to the framework achieved by using \( ^7C_2 \) in this structure.