/*! 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} Problem 10 Eight rooks are placed randomly ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Eight rooks are placed randomly on a chess board (with at most one on each square). What is the probability that: (a) They are all in a straight line? (b) No two are in the same row or column?

Short Answer

Expert verified
(a) Probability is nearly 0; (b) Probability is very small but not zero.

Step by step solution

01

Calculate Total Arrangements

A chess board has 64 squares, and we want to place 8 rooks on it. We first calculate the total number of ways to position them. For the first rook, we have 64 options, 63 options for the second, continuing down to 57 options for the eighth. This is a permutation problem:\[64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57\]
02

Probability Rooks are in a Line

For the rooks to all be in a straight line, they must all be on the same row or same column. There are 8 rows or columns where the rooks can line up. For any one line (row or column), there are 8 possible positions out of 64, which can be selected in:\[\binom{8}{8} = 1\]Thus, the probability they all line up in any particular row or column is:\[\frac{8 \cdot 8!}{64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57}\]
03

Calculate Arrangements with No Two in Same Row or Column

If no two rooks are in the same row or column, we must select 8 different rows and 8 different columns, and place one rook at each intersection. The number of such arrangements is the number of permutations of 8 objects:\[8!\]
04

Probability of No Two Rooks in Same Row or Column

Finally, we find the probability by dividing the favorable outcomes (step 3) by the total arrangements (step 1):\[\frac{8!}{64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57}\]
05

Simplify and Compute

Calculate the numeric probabilities from step 2 and step 4 using the above expressions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Permutations
Permutations are crucial to understanding how arrangements work, especially in problems where order matters. When dealing with permutations, we choose items and organize them in specific sequences. A classic formula for permutations is to multiply descending numbers, starting with the total number of items and selecting down to how many you choose. In the problem, placing 8 rooks on a chessboard involves choosing positions. We begin with 64 options for the first rook, then choose 63 for the next, and so on, reducing choices one by one until the eighth rook. This gives us a large product:
  • 64 choices for the first rook
  • 63 choices for the second rook
  • And so forth, down to 57 for the eighth
Then, the formula becomes:\[64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57\]This product shows all the possible ways 8 rooks can be placed uniquely in sequence across a chessboard. Understanding permutations helps us calculate configurations wherever order is a key part of the problem.
Combinatorial Probability
Combinatorial probability deals with the likelihood of specific arrangements occurring from a set of possibilities. In this exercise, we calculate the chance of specific setups of rooks on a chessboard.
First, consider the probability that all 8 rooks are in a single line, either row or column. As there are 8 rows and 8 columns, there are only 16 possible configurations out of a vastly larger total arrangement.
To determine these probabilities, we find the ratio of desired outcomes (all rooks in one line) to total outcomes (all possible placements). Using permutations, we describe how all 64 squares can receive a rook uniquely:
  • The probability that all rooks are in a single row or column is formulated as:
    \[ \frac{8 \cdot 8!}{64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57} \]
Combinatorial probability simplifies complex scenarios by focusing on desired outcomes and comparing them to all possible outcomes.
Chessboard Problems
Chessboard problems often involve understanding the unique constraints and arrangements possible on an 8x8 grid. The exercise looks at rooks' positions, focusing on criteria like no two rooks sharing the same row or column.
For such placements, leaving rooks to share rows or columns violates the rules of standard non-capturing arrangements. This requires finding and calculating arrangements where no line conflicts occur. We determined that for 8 rooks:
  • We choose 8 distinct rows
  • We choose 8 distinct columns
  • Each rook is placed at the intersection of a unique row and column
This becomes a permutation of 8 items, written as:\[8!\]Then, finding probabilities needs dividing these non-conflicting arrangements by total arrangements calculated with: \[\frac{8!}{64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57}\]Chessboard challenges push problem-solvers to merge strategic understanding of the grid with mathematical techniques like permutations and combinations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that \(N\) objects are placed in a row. The operation \(S_{k}\) is defined as follows: "Pick one of the first \(k\) objects at random and swap it with the object in the \(k\) th place." Now perform \(S_{N}, S_{N-1}, \ldots, S_{1}\). Show that the final arrangement is equally likely to be any one of the \(N\) ! permutations of the objects.

You pick an integer at random between zero and \(10^{5}\) inclusive. What is the probability that its digits are all different?

Tennis Suppose that \(2 n\) players enter for two consecutive tennis tournaments. If the draws for each tournament are random, what is the probability that no two players meet in the first round of both tournaments? If \(n\) is large, show that this probability is about \(e^{-\frac{1}{2}}\).

You have two pairs of red socks, three pairs of mauve socks, and four pairs with a rather attractive rainbow motif. If you pick two socks at random. what is the probability that they match?

A chandelier has seven light bulbs arranged around the circumference of a circle. By the end of a given year, each will have burnt out with probability \(\frac{1}{2}\). Assuming that they do so independently, what is the probability that four or more bulbs will have burnt out? If three bulbs burn out, what is the probability that no two are adjacent? I decide that I will replace all the dead bulbs at the end of the year only if at least two are adjacent. Find the probability that this will happen. If it does, what is the probability that I will need more than two bulbs?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.