Problem 96
Question
A table has provision for 7 seats, 4 being on one side facing the window and 3 being on the opposite side. The number of ways in which 7 people can be seated at the table if 2 people, \(\mathrm{X}\) and \(\mathrm{Y}\), must sit on the same side, is (A) 3260 (B) 2160 (C) 3350 (D) None of these
Step-by-Step Solution
Verified Answer
(D) None of these; the total ways are 1080.
1Step 1: Understanding Seating Arrangement
The total number of seats at the table is 7, with 4 seats on one side and 3 seats on the opposite side. Two specific people, \(\mathrm{X}\) and \(\mathrm{Y}\), must sit on the same side.
2Step 2: Choosing a Side for X and Y
Since X and Y need to sit on the same side, they can either sit on the side with 4 seats or the side with 3 seats. This gives us 2 choices for selecting the side for X and Y.
3Step 3: Calculate Seating Arrangement for X and Y
**Option 1**: If X and Y are on the side with 4 seats, choose 2 seats from the 4 available seats. This can be done in \( \binom{4}{2} = 6 \) ways.**Option 2**: If X and Y are on the side with 3 seats, choose 2 seats from the 3 available seats. This can be done in \( \binom{3}{2} = 3 \) ways.
4Step 4: Calculate Remaining Seats Arrangements
**Option 1**: If X and Y occupy 2 seats on the side with 4 seats, 2 seats remain on that side and 3 seats remain on the opposite side. There are 5 people left to be seated, who can be arranged in these 5 seats in \(5! = 120\) different ways.**Option 2**: If X and Y occupy 2 seats on the side with 3 seats, 1 seat remains on that side and 4 seats remain on the opposite side. Again, 5 people left to be arranged in \(5! = 120\) different ways.
5Step 5: Combine Arrangements
For both options, multiply the choices made for seating X and Y with the arrangements of the remaining seats:- **Option 1**: \(6 \times 120 = 720\)- **Option 2**: \(3 \times 120 = 360\)Add the results from both options to get the total number of possibilities: \(720 + 360 = 1080\).
Key Concepts
CombinatoricsPermutationsBinomial Coefficient
Combinatorics
Combinatorics is a field of mathematics dedicated to counting, arranging, and analyzing ways to combine items, especially when certain arrangements or rules must be followed. When we tackle problems like seating arrangements, we often employ several basic combinatorial techniques.
In the context of the seating arrangement problem, combinatorics helps us count the ways people can sit, considering conditions such as specific people sitting on the same side. By breaking down the problem into steps, we can apply simple counting principles to get a grasp of the solution.
- **Counting Principles**: Understand how many different ways there are to make a set of arrangements.
- **Combinations and Permutations**: Select items in particular orders or without regard to order.
- **Complex Arrangements**: Use rules or restrictions, like in seating arrangements where certain people must sit together.
In the context of the seating arrangement problem, combinatorics helps us count the ways people can sit, considering conditions such as specific people sitting on the same side. By breaking down the problem into steps, we can apply simple counting principles to get a grasp of the solution.
Permutations
Permutations focus on arranging a set of items where the order does matter. When you're arranging seats at a table, the position of each person is important, making permutations a crucial concept.
To calculate permutations, you often use factorial notation, denoted with an exclamation mark \(!\). For example, if you have 5 people competing for 5 seats, the number of possible arrangements can be calculated as \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). This idea was used in our solution when arranging the remaining 5 unseated people in the available seats.
Understanding permutations helps you see how different arrangements yield different seating outcomes. This is essential when figuring out the total number of valid seating arrangements under the problem's constraints.
To calculate permutations, you often use factorial notation, denoted with an exclamation mark \(!\). For example, if you have 5 people competing for 5 seats, the number of possible arrangements can be calculated as \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). This idea was used in our solution when arranging the remaining 5 unseated people in the available seats.
- **Factorials**: Represent the number of ways to arrange a certain number of distinct items.
- **Order Matters**: Permutations require the specific sequence or order of elements.
- **Seating Problem**: Applied to determine different feasible seating options when X and Y's positions were fixed.
Understanding permutations helps you see how different arrangements yield different seating outcomes. This is essential when figuring out the total number of valid seating arrangements under the problem's constraints.
Binomial Coefficient
A binomial coefficient, written as \(\binom{n}{k}\), determines how many ways you can select \(k\) items from a set of \(n\) items, without regard to the order. This concept is pivotal in problems where you have choices or groups to form.
For example, in this seating problem, the binomial coefficient appeared when figuring out how many ways we could seat people X and Y on one side of the table. Let's represent it explicitly:
Employing binomial coefficients allows us to calculate specific choices and combinations quickly. They reduce more complex counting problems into simpler steps by focusing on selection without the need to worry about the order of selection.
For example, in this seating problem, the binomial coefficient appeared when figuring out how many ways we could seat people X and Y on one side of the table. Let's represent it explicitly:
- **\(\binom{4}{2}\)**: Represents the selection of 2 seats from 4, determining the arrangement possibilities for X and Y on one side.
- **\(\binom{3}{2}\)**: Used to decide their placement on the other side, with only 3 seats.
Employing binomial coefficients allows us to calculate specific choices and combinations quickly. They reduce more complex counting problems into simpler steps by focusing on selection without the need to worry about the order of selection.
Other exercises in this chapter
Problem 94
The number of four digit numbers that can be formed from the digits \(0,1,2,3,4,5\) with at least one digit repeated is (A) 420 (B) 560 (C) 780 (D) None of thes
View solution Problem 95
The number of odd numbers lying between 40000 and 70000 that can be made from the digits \(0,1,2,4,5,7\) if digits can be repeated in the same number is (A) 864
View solution Problem 97
There are four oranges, five apples and six mangoes in a fruit basket. The number of ways in which a person can make a selection of fruits among the fruits in t
View solution Problem 98
The number of zeros at the end of \(100 !\) is (A) 36 (B) 18 (C) 24 (D) None of these
View solution