- #1
chwala
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- Homework Statement
- solve the problem below;
- Relevant Equations
- permutation and combination
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Permutation problem: Seven friends queue up for a buffet....
- Thread starter chwala
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- Friends Permutation Queue
In summary, the conversation discusses two different approaches to arranging a group of people into a car with a capacity of 4 or 3 occupants. The first approach involves merging the last two people and arranging the remaining 6 people in 6C4 ways, resulting in 15 possible arrangements. The second approach involves considering the two fellows boarding the car as a group, resulting in 5C2 ways for the remaining 3 people to be arranged, or 5C4 ways for the two fellows and one other person to be arranged, for a total of 15 possible arrangements. The speaker agrees with the second approach.
- #2
chwala
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for the first part, not difficult,
we have ##7!=5040##
now for the second part, i get a bit confused here ok, i merged the last two fellows and now i have 6 items...
therefore i will have ##6C4 ##× the remaining ##3## can be arranged in 1 way only= ##15## is this the correct approach?...not one of my favorite topics
...
or can i say that, the car can be filled up in this way,
##5C2 ×1## way only(remaining 3 people), assuming that the two fellows board the 4-vehicle capacity or
##5C4 ×1 ##way only(2 fellows plus 1 person) , assuming that the two fellows board the vehicle holding 3 occupants...which gives
##10+5=15##
we have ##7!=5040##
now for the second part, i get a bit confused here ok, i merged the last two fellows and now i have 6 items...
therefore i will have ##6C4 ##× the remaining ##3## can be arranged in 1 way only= ##15## is this the correct approach?...not one of my favorite topics
...or can i say that, the car can be filled up in this way,
##5C2 ×1## way only(remaining 3 people), assuming that the two fellows board the 4-vehicle capacity or
##5C4 ×1 ##way only(2 fellows plus 1 person) , assuming that the two fellows board the vehicle holding 3 occupants...which gives
##10+5=15##
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- #3
Delta2
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I agree with your solution.
What is a permutation problem?
A permutation problem is a type of mathematical problem that involves arranging a set of objects or elements in a specific order or sequence. The order of the elements in a permutation is important and can affect the outcome of the problem.
How many possible permutations are there for seven friends queueing up for a buffet?
There are 7! (seven factorial) possible permutations for seven friends queueing up for a buffet. This means that there are 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040 possible ways for the friends to queue up.
What is the formula for calculating permutations?
The formula for calculating permutations is n! / (n-r)! where n is the total number of objects and r is the number of objects being selected. In the case of the seven friends queueing up for a buffet, n = 7 and r = 7, so the formula becomes 7! / (7-7)! = 7! / 0! = 7! = 5040.
What is the difference between a permutation and a combination?
A permutation involves arranging a set of objects in a specific order, while a combination involves selecting a subset of objects from a larger set without regard to order. In the context of the seven friends queueing up for a buffet, a permutation would be the different ways they can stand in line, while a combination would be the different groups of friends that can be formed from the larger group of seven.
How can permutations be used in real-life situations?
Permutations can be used to solve a variety of real-life problems, such as arranging seating at a dinner party, creating unique passwords, or determining the number of possible outcomes in a game or competition. They can also be used in fields such as genetics, computer science, and statistics.
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