Problem 4
Question
If eight persons are to address a meeting then the number of ways in which a specified speaker is to speak before another specified speaker, is (A) 40320 (B) 2520 (C) 20160 (D) None of these
Step-by-Step Solution
Verified Answer
The answer is 20160.
1Step 1: Identify the Problem Type
This is a permutation problem where we are asked to find the number of ways to arrange the order of speakers under a specific condition.
2Step 2: Calculate Total Permutations
Calculate the total number of ways to arrange 8 speakers: 8! (8 factorial), which is given by: \[8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320\]
3Step 3: Assess Condition - Specified Order of Two Speakers
Since only a specified speaker must speak before another specific speaker, without restrictions, each speaker could either speak before or after the other. Therefore, half of the total arrangements (\(\frac{1}{2}\) of 8!) will have the first specified speaker speaking before the second: \[\frac{40320}{2} = 20160\]
4Step 4: Conclude the Result
Thus, the number of ways in which a specified speaker can speak before another specified speaker is 20160.
Key Concepts
Factorial CalculationOrder of ArrangementConditional Permutations
Factorial Calculation
Factorial calculation is a mathematical operation used to determine the number of ways to arrange a set of objects. It is denoted by an exclamation mark, such as in "8!", which reads as "eight factorial". The calculation involves multiplying a series of descending natural numbers. For example, to find 8!, you multiply 8 by every whole number below it:
- Start with 8 and multiply by 7: 8 × 7 = 56
- Multiply the result by 6: 56 × 6 = 336
- Continue this multiplication sequence down to 1
Order of Arrangement
Understanding the order of arrangement is crucial when discussing permutations in mathematics. A permutation is an arrangement of items in a specific sequence or order. If you have a set of items, the focus is not just on which items are included but in what order they appear.
In any scenario where positions matter, permutations come into play. For example, if you have eight speakers and are interested in how they line up to speak, the order they speak in will create different permutations.
In any scenario where positions matter, permutations come into play. For example, if you have eight speakers and are interested in how they line up to speak, the order they speak in will create different permutations.
- The arrangement 'Speaker A, Speaker B, Speaker C' is different from 'Speaker C, Speaker A, Speaker B'.
- Even switching two people changes the entire order, which can be significant.
Conditional Permutations
Conditional permutations take the idea of permutations a step further by introducing specific conditions or rules that must be followed. In the given exercise, the condition states that one specific speaker must speak before another specific speaker.
To handle this requirement, it's important to first calculate all possible arrangements (using the factorial) and then adjust the calculations based on the condition. Here, we note that without any conditions, each person has an equal chance of speaking before or after another.
To handle this requirement, it's important to first calculate all possible arrangements (using the factorial) and then adjust the calculations based on the condition. Here, we note that without any conditions, each person has an equal chance of speaking before or after another.
- The total unconditioned arrangements for 8 people are 8! = 40,320.
- With the condition that one must speak before the other, only half of the arrangements satisfy the requirement.
- Therefore, divide 40,320 by 2, yielding 20,160 conditional permutations.
Other exercises in this chapter
Problem 2
If \(m=\) number of distinct rational numbers \(\frac{p}{q} \in\) \((0,1)\) such that \(p, q \in\\{1,2,3,4,5\\}\) and \(n=\) number of mappings from \(\\{1,2,3\
View solution Problem 3
The letters of the word RANDOM are written in all possible orders and these words are written out as in a dictionary then the rank of the word RANDOM is (A) 614
View solution Problem 5
The number of permutations of letters \(a, b, c, d, e, f, g\) so that neither the pattern beg nor cad appears is (A) \(\frac{7 !}{3 ! 3 !}\) (B) \(\frac{7 !}{2
View solution Problem 6
The number of ways of selecting 10 balls from the unlimited number of red, green, white and yellow balls, if selection must include 2 red and 3 yellow balls, is
View solution