Chapter 9: Problem 87
Prove the identity. \(_{n} P_{n-1}=_{n} P_{n}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 9: Problem 87
Prove the identity. \(_{n} P_{n-1}=_{n} P_{n}\)
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Solve for \(n\) $$4 \cdot_{n+1} P_{2}=_{n+2} P_{3}$$
A U.S. Senate Committee has 14 members. Assuming party affiliation is not a factor in selection, how many different committees are possible from the 100 U.S. senators?
Consider \(n\) independent trials of an experiment in which each trial has two possible outcomes: "success" or "failure." The probability of a success on each trial is \(p,\) and the probability of a failure is \(q=1-p .\) In this context, the term \(_{n} C_{k} p^{k} q^{n-k}\) in the expansion of \((p+q)^{n}\) gives the probability of \(k\) successes in the \(n\) trials of the experiment.To find the probability that the sales representative in Exercise 87 makes four sales when the probability of a sale with any one customer is \(\frac{1}{2},\) evaluate the term $$_{8} C_{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{4}$$, in the expansion of \(\left(\frac{1}{2}+\frac{1}{2}\right)^{8}\).
Use the Binomial Theorem to approximate the quantity accurate to three decimal places. For example, in Exercise \(79,\) use the expansion \(\begin{aligned}(1.02)^{8} &=(1+0.02)^{8} \\ &=1+8(0.02)+28(0.02)^{2}+\cdot \cdot \cdot+(0.02)^{8}\end{aligned}\), $$(1.98)^{9}$$
You are dealt five cards from a standard deck of 52 playing cards. In how many ways can you get (a) a full house and (b) a five-card combination containing two jacks and three aces? (A full house consists of three of one kind and two of another. For example, A-A-A-5-5 and K-K-K-10-10 are full houses.)
What do you think about this solution?
We value your feedback to improve our textbook solutions.