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Chapter 5: Problem 348

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

Expert verified
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

01

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.
02

Rewrite the inequality using the binomial coefficient formula

We can rewrite the inequality using the formula for binomial coefficients: \(\frac{(n-1)!}{5!(n-6)!} + \frac{(n-1)!}{6!(n-7)!} < \frac{n!}{7!(n-8)!}\)
03

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: \((n-1)!7 < n!(n-8)!\)
04

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.

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Most popular questions from this chapter

The number of ordered pairs of integers \((x, y)\) satisfying the equation \(x^{2}+6 x+y^{2}=4\) is (a) 2 (b) 4 (c) 6 (d) 8

10 person are to be arranged around a round table. 3 persons wish to sit as a group number of ways the arrangement can be made is (a) \(9 ! \times 3 !\) (b) \(8 ! \times 3 !\) (c) \(7 ! \times{ }^{8} \mathrm{P}_{3}\) (d) \(7 ! \times 3 !\)

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