Problem 22 - 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) \({ }^{7} P_{2} 2^{5}\) (B) \({ }^{7} C_{2} 2^{5}\) (C) \({ }^{7} C_{2} 5^{2}\) (D) None of these

Step-by-Step Solution

Answer

The answer is (B) \( { }^{7} C_{2} 2^{5} \).

Step 1: Calculate the number of ways to arrange two 2's in seven-digit positions

First, we need to determine how we can position the two '2's in the seven-digit number. Since we need to place exactly two '2's in the number, we use combinations because the order of '2's doesn't matter once their positions are selected. We use the combination formula \( \binom{n}{r} \) where \( n \) is the total number of positions (7) and \( r \) is the number of positions to fill with '2's (2), giving us \( \binom{7}{2} \).

Step 2: Calculate the number of ways to fill the remaining five positions

After placing two '2's, we are left with five positions to fill using the digits '1' and '3'. Each position can be filled with either a '1' or a '3', and since there are 5 such positions, the total number of combinations for these positions is \( 2^5 \), as each of the 5 positions has 2 choices (either '1' or '3').

Step 3: Multiply the results from Step 1 and Step 2

The total number of different 7-digit numbers can be calculated by multiplying the number of ways to position the '2's (from Step 1) by the number of ways to fill the remaining positions (from Step 2). Therefore, the total number of ways is \( \binom{7}{2} \times 2^5 \).

Key Concepts

Permutations

Permutations are an important concept in combinatorics that deal with the ordering of elements. When you have a set of items and you want to know how many ways you can arrange them, permutations are used. Each unique arrangement of the items is called a permutation.
Permutations are concerned with the order of the items, unlike combinations. In technical terms, permutations consider arrangements where the sequence matters. For example, arranging the letters A, B, and C would have permutations like ABC, BAC, CAB, etc. The formula for permutations when selecting all elements is given as:\[ n! = n \times (n-1) \times (n-2) \times \ldots \times 1\]Where \( n \) is the total number of items to arrange. If you want permutations of a subset \( r \) from \( n \), the formula is:\[^nP_r = \frac{n!}{(n-r)!}\]This is useful in situations where each position or sequence counts separately. In our exercise, permutations don't apply directly because the placement of digits '1' and '3' doesn't require sequence consideration after selecting positions for '2's.

Combinations

Combinations are used when you need to select items from a group, but the order doesn't matter. This makes combinations distinct from permutations where sequence is crucial.
In combinations, you're only interested in the subset selection, not the arrangement. For instance, if you're selecting 2 fruits from a basket of apples, oranges, and bananas, AR and RA are considered the same in combinations.
The formula to determine combinations is:\[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]Where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. In the exercise, we use combinations to determine how the two '2's can be arranged among the seven digits. Specifically, we calculate \( \binom{7}{2} \), which tells us the number of ways to place two '2's in seven positions without considering the sequence, only the spots where '2's will be.

Binomial Theorem

The Binomial Theorem provides a powerful way to expand expressions of the form \((x + y)^n\) using algebra. It relates to combinatorics by involving coefficients known as binomial coefficients, which are analogous to combinations.
The general formula of the Binomial Theorem is:\[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\]Each term in this expansion corresponds to a combination calculation. Here, \( \binom{n}{k}\) denotes the binomial coefficient and signifies the number of ways to choose \( k \) occurrences of one term (like \( y \)) in a product of \( n \) such terms.
Although the binomial theorem isn't directly used in solving the exercise, its principles help understand how combinations work, like figuring out \( \binom{7}{2} \) for placing '2's, as these values are core parts of binomial expansions, indicative of the combinatorial properties involved in the exercise.