Question

The least positive integer \(\mathrm{n}\) for which \({ }^{\mathrm{n}-1} \mathrm{C}_{5}+{ }^{\mathrm{n}-1} \mathrm{C}_{6}<{ }^{\mathrm{n}} \mathrm{C}_{7}\) is (a) 14 (b) 15 (c) 16 (d) 28

Short answer

The least positive integer n for which the inequality ${ }^{n-1}C_{5}+{ }^{n-1}C_{6}<{ }^{n}C_{7}$ is satisfied is \(n = 14\).

Step-by-step solution

Understand the binomial coefficients

The binomial coefficients are represented as \({}^{n}C_{r}\), which is read as "n choose r." It represents the number of ways to choose r elements from a set of n elements and is calculated using the formula: \({}^{n}C_{r} = \frac{n!}{r!(n-r)!}\), where n! represents the factorial of n.

Rewrite the inequality using the binomial coefficient formula

We can rewrite the inequality using the formula for binomial coefficients: \(\frac{(n-1)!}{5!(n-6)!} + \frac{(n-1)!}{6!(n-7)!} < \frac{n!}{7!(n-8)!}\)

Simplify the inequality

We can simplify the inequality by eliminating the factorials. \(\frac{(n-1)!}{(n-6)!}(\frac{1}{5!} + \frac{1}{6!}) < \frac{n!}{(n-8)!}\frac{1}{7!}\) Now, multiply both sides by \((n-6)!(n-8)!5!6!7!\) to get rid of the fractions: \((n-1)!(n-8)!6!7! < n!(n-6)!6!5!\) Now, cancel out the common terms: \((n-1)!7 < n!(n-8)!\)

Test the given options

Now, we can test each of the given options to see which one satisfies the inequality. (a) 14: \((14-1)!7 < 14!(14-8)! \implies 13!7 < 14!6! \implies 7 < 14\), which is true. Therefore, the least positive integer n for which the inequality is satisfied is 14, and the correct answer is (a) 14.

Key concepts

Factorial

Factorials are a fundamental concept in combinatorics, often appearing in the calculation of binomial coefficients. A factorial, denoted as \( n! \), represents the product of all positive integers from 1 to \( n \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials are used in permutations and combinations where order matters in arranging elements from a set.

They grow very quickly, and the term 'factorial' originates from their multiplicative nature. One important property of factorials is that \( (n+1)! = (n+1) \times n! \). This recursive property is crucial in simplifying many combinatorics problems and inequalities like in our exercise, allowing for elimination and cancellation of terms in expressions.

Inequality

Inequalities are mathematical statements that describe the relationship between two values, indicating that one value is either larger or smaller than the other. In combinatorics, inequalities are often used to prove properties or find specific conditions where one combination is greater or less than another.

In our exercise, the inequality was presented in terms of binomial coefficients. To solve it, we required algebraic manipulation including simplification by eliminating common factors sustained by the factorials. Effective use of inequalities helps in narrowing down potential values for variables, allowing us to test feasible solutions which satisfy the condition set by the inequality.

Combinatorics

Combinatorics is a branch of mathematics concerning the counting, arrangement, and combination of objects. In combinatorial problems, objects from a set are chosen and organized according to certain rules, leading to calculations such as permutations and combinations.

Binomial coefficients, expressed as \( ^{n}C_{r} \) are at the heart of combinatorics, representing the number of ways a smaller subset can be selected from a larger set. They are calculated using the factorial function and help solve problems related to probability, graph theory, and many other fields.

In the given exercise, combinatorials come into play with determining the "choose r" calculation using binomial coefficients and applying these to satisfy the given inequality. Such problems blend algebraic manipulation with the conceptual understanding of arrangements and combinations.