Problem 19 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \({ }^{n} C_{4},{ }^{n} C_{5}\) and \({ }^{n} C_{6}\) are in A.P., then \(n\) can be : [Jan.12, 2019 (II)] (a) 9 (b) 14 (c) 11 (d) 12

Step-by-Step Solution

Answer

n can be 14.

Step 1: Understanding Combinations

The notation \({ }^{n} C_{r}\) represents combinations, calculated as \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). We have \({ }^{n} C_{4},{ }^{n} C_{5},{ }^{n} C_{6}\) in arithmetic progression (A.P.), which means \(2 \times { }^{n} C_{5} = { }^{n} C_{4} + { }^{n} C_{6}\).

Step 2: Express Combinations Using Factorials

Write the expressions for combinations: \({ }^{n} C_{4} = \frac{n!}{4!(n-4)!}\), \({ }^{n} C_{5} = \frac{n!}{5!(n-5)!}\), and \({ }^{n} C_{6} = \frac{n!}{6!(n-6)!}\).

Step 3: Setting Up the A.P. Condition

Use the A.P. condition: \(2 \times { }^{n} C_{5} = { }^{n} C_{4} + { }^{n} C_{6}\). Substitute the factorial expressions: \(2 \times \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}\).

Step 4: Simplifying the Equation

Rewrite the equation:\[2 \times \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!}.\]Multiply through by \(n!\) to simplify:\[2 \times \frac{n(n-1)(n-2)(n-3)(n-4)}{5!} = \frac{n(n-1)(n-2)(n-3)}{4!} + \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!}.\]

Step 5: Equating and Solving for n

For the common factor \(n(n-1)(n-2)(n-3)\) cancel out, leaving:\[2(n-4) = 5 + \frac{(n-4)(n-5)}{6}.\]Simplify by multiplying through with 6 to eliminate the fraction:\[12(n-4) = 30 + (n-4)(n-5).\]Simplify and solve the quadratic equation:\[12n - 48 = 30 + n^2 - 9n + 20.\]\[n^2 - 21n + 98 = 0.\]

Step 6: Solving the Quadratic Equation

Solve the quadratic \(n^2 - 21n + 98 = 0\) using the quadratic formula:\[n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]For \(a = 1, b = -21, c = 98\), we have:\[n = \frac{21 \pm \sqrt{441 - 392}}{2}.\]\[n = \frac{21 \pm \sqrt{49}}{2}.\]\(n = 14\) or \(n = 7\). Use valid option: \(n = 14\).

Key Concepts

Combinations in Mathematics

In mathematics, combinations refer to the selection of items from a larger set without considering the order. It is represented by the notation \( \binom{n}{r} \), which stands for the number of ways to choose \( r \) elements from a set of \( n \) elements. This is calculated using the formula:

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

The factorial notation \( n! \) means the product of all positive integers up to \( n \), so \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \).
For example, the combination \( \binom{5}{3} \) or choosing 3 elements from 5, is calculated as:

• \( \binom{5}{3} = \frac{5!}{3! \cdot 2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 2 \times 1} = 10 \)

Combinations are widely used in probability and other areas of mathematics where the arrangement of items is irrelevant and the focus is on selection only.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2, and they take the general form:

• \( ax^2 + bx + c = 0 \)

where \( a, b, \) and \( c \) are constants, and \( a eq 0 \).
The solutions to a quadratic equation can be found using different methods such as factoring, completing the square, or using the quadratic formula. The quadratic formula provides the solutions for the equation as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

The part under the square root, \( b^2 - 4ac \), is called the discriminant. It can tell us the nature of the roots:

• If \( b^2 - 4ac > 0 \), the equation has two distinct real roots.
• If \( b^2 - 4ac = 0 \), it has exactly one real root.
• If \( b^2 - 4ac < 0 \), there are no real roots, the roots are complex.

Understanding quadratic equations is essential since they appear in various mathematical problems and real-world situations.

Binomial Coefficients

Binomial coefficients are coefficients in the expansion of a binomial raised to a power, described by the binomial theorem. The binomial theorem states that:

• \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \)

Each term in the expansion is associated with a binomial coefficient \( \binom{n}{k} \).
For instance, for \((x + y)^3\), the expansion would be \(x^3 + 3x^2y + 3xy^2 + y^3\), with coefficients 1, 3, 3, and 1, respectively.

• The binomial coefficient \( \binom{n}{k} \) can be calculated as \( \frac{n!}{k!(n-k)!} \).

These coefficients also appear in combinatorial counting, as they represent the number of ways to select \( k \) elements from \( n \) elements, closely linked with combinations.
Binomial coefficients have widespread applications, including probability, algebra, and calculus. Recognizing their patterns and uses helps solve mathematical problems involving polynomial expansions and combinations.