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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Color the edges and diagonals of a regular polygon

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Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
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