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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

A round table conference is to be held among 20 delegates of 20 countries. The no. of ways they can be seated if two particular delegates are never sit together is. (a) \(17 \cdot 18 !\) (b) \(18 \cdot 19 !\) (c) \((20 ! / 2)\) (d) \(19 ! \cdot 2\)

The number of arrangements that can be made out of the letter of the word " SUCCESS " so that the all S's do not come together is (a) 60 (b) 120 (c) 360 (d) 420

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choice available to him is (a) 140 (b) 196 (c) 180 (d) 346

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) \(7 ! \times 3 !\)

The number of integer \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\), such that \(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=20\) and \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \geq 0\) is (a) \({ }^{24} \mathrm{C}_{3}\) (b) \({ }^{25} \mathrm{C}_{3}\) (c) \({ }^{26} \mathrm{C}_{3}\) (d) \({ }^{27} \mathrm{C}_{3}\)
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