what is the probability that in a random group of seven people two were born on Monday and two on Sunday?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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1
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If people can be born with the same probability any day of the week, what is the probability that in a random group of seven people two were born on Monday and two on Sunday?
My analysis:
7 people can be born in 7^7 ways.
7 people can be shuffled in 7! Ways.
4 people occupy two days and the 3 remaining people occupy the 5 other days in 5^3
My analysis lead to the following solution: 7!*5^3/7^7
Is that correct?




probability




share|cite|improve this question



















  • Here I would guess a concept of partitions can be used. en.wikipedia.org/wiki/Partition_(number_theory)
    – mathreadler
    Jul 14 at 21:48














up vote
1
down vote

favorite












If people can be born with the same probability any day of the week, what is the probability that in a random group of seven people two were born on Monday and two on Sunday?
My analysis:
7 people can be born in 7^7 ways.
7 people can be shuffled in 7! Ways.
4 people occupy two days and the 3 remaining people occupy the 5 other days in 5^3
My analysis lead to the following solution: 7!*5^3/7^7
Is that correct?




probability




share|cite|improve this question



















  • Here I would guess a concept of partitions can be used. en.wikipedia.org/wiki/Partition_(number_theory)
    – mathreadler
    Jul 14 at 21:48












up vote
1
down vote

favorite









up vote
1
down vote

favorite











If people can be born with the same probability any day of the week, what is the probability that in a random group of seven people two were born on Monday and two on Sunday?
My analysis:
7 people can be born in 7^7 ways.
7 people can be shuffled in 7! Ways.
4 people occupy two days and the 3 remaining people occupy the 5 other days in 5^3
My analysis lead to the following solution: 7!*5^3/7^7
Is that correct?




probability




share|cite|improve this question











If people can be born with the same probability any day of the week, what is the probability that in a random group of seven people two were born on Monday and two on Sunday?
My analysis:
7 people can be born in 7^7 ways.
7 people can be shuffled in 7! Ways.
4 people occupy two days and the 3 remaining people occupy the 5 other days in 5^3
My analysis lead to the following solution: 7!*5^3/7^7
Is that correct?





probability





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 14 at 21:26









Rayri

485




485











  • Here I would guess a concept of partitions can be used. en.wikipedia.org/wiki/Partition_(number_theory)
    – mathreadler
    Jul 14 at 21:48
















  • Here I would guess a concept of partitions can be used. en.wikipedia.org/wiki/Partition_(number_theory)
    – mathreadler
    Jul 14 at 21:48















Here I would guess a concept of partitions can be used. en.wikipedia.org/wiki/Partition_(number_theory)
– mathreadler
Jul 14 at 21:48




Here I would guess a concept of partitions can be used. en.wikipedia.org/wiki/Partition_(number_theory)
– mathreadler
Jul 14 at 21:48










3 Answers
3






active

oldest

votes

















up vote
0
down vote



accepted










I will try to make an explanation for Phil:s answer.



  1. We first choose the four people to have birthday on either sunday or monday. This can be done with something called "combinations" sometimes written nCr or $^nC_r$. 4 among 7 can be chosen in $^7C_4$ ways.

  2. Then second factor is also an nCr, it is when whe calculate how many ways we can split those 4 people into groups of 2. One group for sunday and one for monday.

  3. Last we have 3 people left and they can be put in any of the remaining free 5 days and that's where the $5^3=125$ comes from.





share|cite|improve this answer

















  • 1




    Thanks for explaining my answer. For your item 2, there are 4! ways to arrange 4 people, but because we have 2 pairs of the same day, we reduce this amount by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:19










  • You are both amazing tutors. Thanks you a lot
    – Rayri
    Jul 14 at 22:23

















up vote
1
down vote













There are only $^7C_4cdot frac4!2!cdot 2!cdot 5^3$ ways to have four out of seven people with two birthdays on Monday and two on Tuesday.



$P(textmmtt) = frac^7C_4cdot frac4!2!cdot 2!cdot 5^37^7 = .03187$






share|cite|improve this answer





















  • Can you explain the procedure of the numerator. how did you apply the multiplication rule?
    – Rayri
    Jul 14 at 22:04











  • This looks more reasonable, but maybe you can explain the different factors and why they are there to those who are new to combinatorics?
    – mathreadler
    Jul 14 at 22:07










  • I thought 7! Was to shuffle the order of the 7 people but now you introduced the idea of 4!/2!*2! Can you please elaborate more on it?
    – Rayri
    Jul 14 at 22:20











  • Yes, $frac4!2!cdot 2!$ where $4!$ is the number of ways you can rearrange 4 people, but because there are 2 pairs having the same birth day, we need to reduce this by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:23











  • @Rayen Sorry I wasn't around to answer your questions earlier, I hope you got your answers, let me know if you have other questions.
    – Phil H
    Jul 14 at 22:29

















up vote
0
down vote













First we choose 2 people among 7 that are born on monday and then we choose 2 people among 5 that are born on Sunday. So the number of good outcomes is $$7choose 2cdot 5choose 2 cdot 5^3= 7cdot 3cdot 5cdot 2cdot 5^3 $$
so the probability you seek is $$P = 30cdot 5^3over 7^6$$






share|cite|improve this answer























  • However, I checked the solution and it was: 7!*5^3/24*7^7
    – Rayri
    Jul 14 at 21:54










  • If we ignore sunday and monday and 5 days left to distribute 3 people over, that's $5^3=125$ which is already bigger than $30$
    – mathreadler
    Jul 14 at 21:55











  • @Rayen then why did you click accept answer?
    – mathreadler
    Jul 14 at 21:55










  • Because in fact there is nothing wrong with your answer. Thanks you a lot ❤️
    – Rayri
    Jul 14 at 22:01










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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










I will try to make an explanation for Phil:s answer.



  1. We first choose the four people to have birthday on either sunday or monday. This can be done with something called "combinations" sometimes written nCr or $^nC_r$. 4 among 7 can be chosen in $^7C_4$ ways.

  2. Then second factor is also an nCr, it is when whe calculate how many ways we can split those 4 people into groups of 2. One group for sunday and one for monday.

  3. Last we have 3 people left and they can be put in any of the remaining free 5 days and that's where the $5^3=125$ comes from.





share|cite|improve this answer

















  • 1




    Thanks for explaining my answer. For your item 2, there are 4! ways to arrange 4 people, but because we have 2 pairs of the same day, we reduce this amount by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:19










  • You are both amazing tutors. Thanks you a lot
    – Rayri
    Jul 14 at 22:23














up vote
0
down vote



accepted










I will try to make an explanation for Phil:s answer.



  1. We first choose the four people to have birthday on either sunday or monday. This can be done with something called "combinations" sometimes written nCr or $^nC_r$. 4 among 7 can be chosen in $^7C_4$ ways.

  2. Then second factor is also an nCr, it is when whe calculate how many ways we can split those 4 people into groups of 2. One group for sunday and one for monday.

  3. Last we have 3 people left and they can be put in any of the remaining free 5 days and that's where the $5^3=125$ comes from.





share|cite|improve this answer

















  • 1




    Thanks for explaining my answer. For your item 2, there are 4! ways to arrange 4 people, but because we have 2 pairs of the same day, we reduce this amount by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:19










  • You are both amazing tutors. Thanks you a lot
    – Rayri
    Jul 14 at 22:23












up vote
0
down vote



accepted







up vote
0
down vote



accepted






I will try to make an explanation for Phil:s answer.



  1. We first choose the four people to have birthday on either sunday or monday. This can be done with something called "combinations" sometimes written nCr or $^nC_r$. 4 among 7 can be chosen in $^7C_4$ ways.

  2. Then second factor is also an nCr, it is when whe calculate how many ways we can split those 4 people into groups of 2. One group for sunday and one for monday.

  3. Last we have 3 people left and they can be put in any of the remaining free 5 days and that's where the $5^3=125$ comes from.





share|cite|improve this answer













I will try to make an explanation for Phil:s answer.



  1. We first choose the four people to have birthday on either sunday or monday. This can be done with something called "combinations" sometimes written nCr or $^nC_r$. 4 among 7 can be chosen in $^7C_4$ ways.

  2. Then second factor is also an nCr, it is when whe calculate how many ways we can split those 4 people into groups of 2. One group for sunday and one for monday.

  3. Last we have 3 people left and they can be put in any of the remaining free 5 days and that's where the $5^3=125$ comes from.






share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 14 at 22:14









mathreadler

13.6k71857




13.6k71857







  • 1




    Thanks for explaining my answer. For your item 2, there are 4! ways to arrange 4 people, but because we have 2 pairs of the same day, we reduce this amount by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:19










  • You are both amazing tutors. Thanks you a lot
    – Rayri
    Jul 14 at 22:23












  • 1




    Thanks for explaining my answer. For your item 2, there are 4! ways to arrange 4 people, but because we have 2 pairs of the same day, we reduce this amount by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:19










  • You are both amazing tutors. Thanks you a lot
    – Rayri
    Jul 14 at 22:23







1




1




Thanks for explaining my answer. For your item 2, there are 4! ways to arrange 4 people, but because we have 2 pairs of the same day, we reduce this amount by dividing by 2! twice.
– Phil H
Jul 14 at 22:19




Thanks for explaining my answer. For your item 2, there are 4! ways to arrange 4 people, but because we have 2 pairs of the same day, we reduce this amount by dividing by 2! twice.
– Phil H
Jul 14 at 22:19












You are both amazing tutors. Thanks you a lot
– Rayri
Jul 14 at 22:23




You are both amazing tutors. Thanks you a lot
– Rayri
Jul 14 at 22:23










up vote
1
down vote













There are only $^7C_4cdot frac4!2!cdot 2!cdot 5^3$ ways to have four out of seven people with two birthdays on Monday and two on Tuesday.



$P(textmmtt) = frac^7C_4cdot frac4!2!cdot 2!cdot 5^37^7 = .03187$






share|cite|improve this answer





















  • Can you explain the procedure of the numerator. how did you apply the multiplication rule?
    – Rayri
    Jul 14 at 22:04











  • This looks more reasonable, but maybe you can explain the different factors and why they are there to those who are new to combinatorics?
    – mathreadler
    Jul 14 at 22:07










  • I thought 7! Was to shuffle the order of the 7 people but now you introduced the idea of 4!/2!*2! Can you please elaborate more on it?
    – Rayri
    Jul 14 at 22:20











  • Yes, $frac4!2!cdot 2!$ where $4!$ is the number of ways you can rearrange 4 people, but because there are 2 pairs having the same birth day, we need to reduce this by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:23











  • @Rayen Sorry I wasn't around to answer your questions earlier, I hope you got your answers, let me know if you have other questions.
    – Phil H
    Jul 14 at 22:29














up vote
1
down vote













There are only $^7C_4cdot frac4!2!cdot 2!cdot 5^3$ ways to have four out of seven people with two birthdays on Monday and two on Tuesday.



$P(textmmtt) = frac^7C_4cdot frac4!2!cdot 2!cdot 5^37^7 = .03187$






share|cite|improve this answer





















  • Can you explain the procedure of the numerator. how did you apply the multiplication rule?
    – Rayri
    Jul 14 at 22:04











  • This looks more reasonable, but maybe you can explain the different factors and why they are there to those who are new to combinatorics?
    – mathreadler
    Jul 14 at 22:07










  • I thought 7! Was to shuffle the order of the 7 people but now you introduced the idea of 4!/2!*2! Can you please elaborate more on it?
    – Rayri
    Jul 14 at 22:20











  • Yes, $frac4!2!cdot 2!$ where $4!$ is the number of ways you can rearrange 4 people, but because there are 2 pairs having the same birth day, we need to reduce this by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:23











  • @Rayen Sorry I wasn't around to answer your questions earlier, I hope you got your answers, let me know if you have other questions.
    – Phil H
    Jul 14 at 22:29












up vote
1
down vote










up vote
1
down vote









There are only $^7C_4cdot frac4!2!cdot 2!cdot 5^3$ ways to have four out of seven people with two birthdays on Monday and two on Tuesday.



$P(textmmtt) = frac^7C_4cdot frac4!2!cdot 2!cdot 5^37^7 = .03187$






share|cite|improve this answer













There are only $^7C_4cdot frac4!2!cdot 2!cdot 5^3$ ways to have four out of seven people with two birthdays on Monday and two on Tuesday.



$P(textmmtt) = frac^7C_4cdot frac4!2!cdot 2!cdot 5^37^7 = .03187$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 14 at 21:57









Phil H

1,8442311




1,8442311











  • Can you explain the procedure of the numerator. how did you apply the multiplication rule?
    – Rayri
    Jul 14 at 22:04











  • This looks more reasonable, but maybe you can explain the different factors and why they are there to those who are new to combinatorics?
    – mathreadler
    Jul 14 at 22:07










  • I thought 7! Was to shuffle the order of the 7 people but now you introduced the idea of 4!/2!*2! Can you please elaborate more on it?
    – Rayri
    Jul 14 at 22:20











  • Yes, $frac4!2!cdot 2!$ where $4!$ is the number of ways you can rearrange 4 people, but because there are 2 pairs having the same birth day, we need to reduce this by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:23











  • @Rayen Sorry I wasn't around to answer your questions earlier, I hope you got your answers, let me know if you have other questions.
    – Phil H
    Jul 14 at 22:29
















  • Can you explain the procedure of the numerator. how did you apply the multiplication rule?
    – Rayri
    Jul 14 at 22:04











  • This looks more reasonable, but maybe you can explain the different factors and why they are there to those who are new to combinatorics?
    – mathreadler
    Jul 14 at 22:07










  • I thought 7! Was to shuffle the order of the 7 people but now you introduced the idea of 4!/2!*2! Can you please elaborate more on it?
    – Rayri
    Jul 14 at 22:20











  • Yes, $frac4!2!cdot 2!$ where $4!$ is the number of ways you can rearrange 4 people, but because there are 2 pairs having the same birth day, we need to reduce this by dividing by 2! twice.
    – Phil H
    Jul 14 at 22:23











  • @Rayen Sorry I wasn't around to answer your questions earlier, I hope you got your answers, let me know if you have other questions.
    – Phil H
    Jul 14 at 22:29















Can you explain the procedure of the numerator. how did you apply the multiplication rule?
– Rayri
Jul 14 at 22:04





Can you explain the procedure of the numerator. how did you apply the multiplication rule?
– Rayri
Jul 14 at 22:04













This looks more reasonable, but maybe you can explain the different factors and why they are there to those who are new to combinatorics?
– mathreadler
Jul 14 at 22:07




This looks more reasonable, but maybe you can explain the different factors and why they are there to those who are new to combinatorics?
– mathreadler
Jul 14 at 22:07












I thought 7! Was to shuffle the order of the 7 people but now you introduced the idea of 4!/2!*2! Can you please elaborate more on it?
– Rayri
Jul 14 at 22:20





I thought 7! Was to shuffle the order of the 7 people but now you introduced the idea of 4!/2!*2! Can you please elaborate more on it?
– Rayri
Jul 14 at 22:20













Yes, $frac4!2!cdot 2!$ where $4!$ is the number of ways you can rearrange 4 people, but because there are 2 pairs having the same birth day, we need to reduce this by dividing by 2! twice.
– Phil H
Jul 14 at 22:23





Yes, $frac4!2!cdot 2!$ where $4!$ is the number of ways you can rearrange 4 people, but because there are 2 pairs having the same birth day, we need to reduce this by dividing by 2! twice.
– Phil H
Jul 14 at 22:23













@Rayen Sorry I wasn't around to answer your questions earlier, I hope you got your answers, let me know if you have other questions.
– Phil H
Jul 14 at 22:29




@Rayen Sorry I wasn't around to answer your questions earlier, I hope you got your answers, let me know if you have other questions.
– Phil H
Jul 14 at 22:29










up vote
0
down vote













First we choose 2 people among 7 that are born on monday and then we choose 2 people among 5 that are born on Sunday. So the number of good outcomes is $$7choose 2cdot 5choose 2 cdot 5^3= 7cdot 3cdot 5cdot 2cdot 5^3 $$
so the probability you seek is $$P = 30cdot 5^3over 7^6$$






share|cite|improve this answer























  • However, I checked the solution and it was: 7!*5^3/24*7^7
    – Rayri
    Jul 14 at 21:54










  • If we ignore sunday and monday and 5 days left to distribute 3 people over, that's $5^3=125$ which is already bigger than $30$
    – mathreadler
    Jul 14 at 21:55











  • @Rayen then why did you click accept answer?
    – mathreadler
    Jul 14 at 21:55










  • Because in fact there is nothing wrong with your answer. Thanks you a lot ❤️
    – Rayri
    Jul 14 at 22:01














up vote
0
down vote













First we choose 2 people among 7 that are born on monday and then we choose 2 people among 5 that are born on Sunday. So the number of good outcomes is $$7choose 2cdot 5choose 2 cdot 5^3= 7cdot 3cdot 5cdot 2cdot 5^3 $$
so the probability you seek is $$P = 30cdot 5^3over 7^6$$






share|cite|improve this answer























  • However, I checked the solution and it was: 7!*5^3/24*7^7
    – Rayri
    Jul 14 at 21:54










  • If we ignore sunday and monday and 5 days left to distribute 3 people over, that's $5^3=125$ which is already bigger than $30$
    – mathreadler
    Jul 14 at 21:55











  • @Rayen then why did you click accept answer?
    – mathreadler
    Jul 14 at 21:55










  • Because in fact there is nothing wrong with your answer. Thanks you a lot ❤️
    – Rayri
    Jul 14 at 22:01












up vote
0
down vote










up vote
0
down vote









First we choose 2 people among 7 that are born on monday and then we choose 2 people among 5 that are born on Sunday. So the number of good outcomes is $$7choose 2cdot 5choose 2 cdot 5^3= 7cdot 3cdot 5cdot 2cdot 5^3 $$
so the probability you seek is $$P = 30cdot 5^3over 7^6$$






share|cite|improve this answer















First we choose 2 people among 7 that are born on monday and then we choose 2 people among 5 that are born on Sunday. So the number of good outcomes is $$7choose 2cdot 5choose 2 cdot 5^3= 7cdot 3cdot 5cdot 2cdot 5^3 $$
so the probability you seek is $$P = 30cdot 5^3over 7^6$$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 14 at 22:23


























answered Jul 14 at 21:47









greedoid

26.6k93574




26.6k93574











  • However, I checked the solution and it was: 7!*5^3/24*7^7
    – Rayri
    Jul 14 at 21:54










  • If we ignore sunday and monday and 5 days left to distribute 3 people over, that's $5^3=125$ which is already bigger than $30$
    – mathreadler
    Jul 14 at 21:55











  • @Rayen then why did you click accept answer?
    – mathreadler
    Jul 14 at 21:55










  • Because in fact there is nothing wrong with your answer. Thanks you a lot ❤️
    – Rayri
    Jul 14 at 22:01
















  • However, I checked the solution and it was: 7!*5^3/24*7^7
    – Rayri
    Jul 14 at 21:54










  • If we ignore sunday and monday and 5 days left to distribute 3 people over, that's $5^3=125$ which is already bigger than $30$
    – mathreadler
    Jul 14 at 21:55











  • @Rayen then why did you click accept answer?
    – mathreadler
    Jul 14 at 21:55










  • Because in fact there is nothing wrong with your answer. Thanks you a lot ❤️
    – Rayri
    Jul 14 at 22:01















However, I checked the solution and it was: 7!*5^3/24*7^7
– Rayri
Jul 14 at 21:54




However, I checked the solution and it was: 7!*5^3/24*7^7
– Rayri
Jul 14 at 21:54












If we ignore sunday and monday and 5 days left to distribute 3 people over, that's $5^3=125$ which is already bigger than $30$
– mathreadler
Jul 14 at 21:55





If we ignore sunday and monday and 5 days left to distribute 3 people over, that's $5^3=125$ which is already bigger than $30$
– mathreadler
Jul 14 at 21:55













@Rayen then why did you click accept answer?
– mathreadler
Jul 14 at 21:55




@Rayen then why did you click accept answer?
– mathreadler
Jul 14 at 21:55












Because in fact there is nothing wrong with your answer. Thanks you a lot ❤️
– Rayri
Jul 14 at 22:01




Because in fact there is nothing wrong with your answer. Thanks you a lot ❤️
– Rayri
Jul 14 at 22:01












 

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What is the equation of a 3D cone with generalised tilt?

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Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an
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