Problem 154 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \({ }^{n} C_{r}\) denotes the number of combinations of \(n\) things taken \(r\) at a time, then the expression \({ }^{n} C_{r+1}+{ }^{n} C_{r-1}+\) \(2 \times{ }^{n} C_{r}\) equals (A) \({ }^{n+2} C_{r}\) (B) \({ }^{n+2} C_{r+1}\) (C) \({ }^{n+1} C_{r}\) (D) \({ }^{n+1} C_{r+1}\)

Step-by-Step Solution

Answer

The expression equals \({ }^{n+2} C_{r+1}\), option (B).

Step 1: Understanding the Combination Formula

First, recall that the combination formula for choosing \(r\) items from \(n\) is given by \({ }^{n} C_{r} = \frac{n!}{r!(n-r)!}\). This represents how many ways you can choose \(r\) items from \(n\) without regard to order.

Step 2: Apply the Pascal's Identity

Use Pascal's Identity, which states that \({ }^{n} C_{r} + { }^{n} C_{r-1} = { }^{n+1} C_{r}\). We know \(2 \times { }^{n} C_{r} = { }^{n} C_{r} + { }^{n} C_{r}\) to incorporate into this identity.

Step 3: Combine Using the Identity

First calculate \({ }^{n} C_{r+1} + { }^{n} C_{r-1}\) using Pascal's identity: \({ }^{n} C_{r+1} + { }^{n} C_{r-1} = { }^{n+1} C_{r}\). Now add \(2 \times { }^{n} C_{r}\) yielding \({ }^{n+1} C_{r} + { }^{n} C_{r}\).

Step 4: Simplification using Combination Properties

Recognize that \({ }^{n+1} C_{r} + { }^{n} C_{r} = { }^{n+2} C_{r+1}\) using the expanded Pascal's Identity. Thus, combining all parts, we have: \({ }^{n} C_{r+1} + { }^{n} C_{r-1} + 2 \times { }^{n} C_{r} = { }^{n+2} C_{r+1}\).

Step 5: Conclusion

The original expression \({ }^{n} C_{r+1} + { }^{n} C_{r-1} + 2 \times { }^{n} C_{r} = { }^{n+2} C_{r+1}\), which matches option \(B\).

Key Concepts

Pascal's Identity

Pascal's Identity is a crucial concept in combinatorics. It's like a special rule that helps simplify and solve problems involving combinations. The identity states that the sum of two combinations, specifically \( { }^{n} C_{r} + { }^{n} C_{r-1} \), equals another combination: \( { }^{n+1} C_{r} \).
This might sound a bit abstract, but think of it like adding counts of ways to choose without changing the overall selection.
When you are working with combinations, if you need to find out how many ways there are to choose certain objects, Pascal's Identity can help. It allows you to combine calculations and figure out solutions quickly.
For the given problem, using Pascal's Identity, we combined different subsections of the original equation, like \( { }^{n} C_{r+1} \), which helped in moving step by step towards simplifying the expression.

Combination Formula

The Combination Formula is a foundational tool for calculating how many ways we can select a group of items from a larger pool. Without caring about the order of selection, the formula is given by: \( { }^{n} C_{r} = \frac{n!}{r!(n-r)!} \). This expression employs factorials, which are the products of all numbers up to a certain number.
Understanding this formula is essential when tackling problems involving combinations of items.

• \( n! \) ("n factorial") is the product from \( n \) down to 1.
• \( r! \) is similarly calculated for \( r \).
• \( (n-r)! \) covers the remaining objects not chosen.

Using this formula, you can calculate specific cases, like \( { }^{5} C_{3} \), which represents choosing 3 items from 5. This becomes crucial in simplifying complex expressions like the one in the exercise.

Binomial Theorem

The Binomial Theorem is a powerful tool used for expanding expressions raised to a power. It states that: \((x + y)^n\) can be expressed as a sum involving terms of the form \(a*b\), where

• a is a binomial coefficient \({ }^{n} C_{r}\).
• b is the result of a combination of powers from x and y.

This theorem ties closely with combinations as the coefficients in the expanded polynomial are exactly those computed by combination formulas.
When applied within complex mathematical problems, like in the exercise, several different combinations can be simplified using the theorem. It helps to see complicated algebraic expressions in a simplified, more solvable way.
In the exercise, understanding combination relations and expansions assists in combining and simplifying to arrive at the desired result.