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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?

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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.
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    $\begingroup$ 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. $\endgroup$
    – Phil H
    Jul 14 '18 at 22:19
  • $\begingroup$ You are both amazing tutors. Thanks you a lot $\endgroup$
    – Rayri
    Jul 14 '18 at 22:23
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There are only $^7C_4\cdot \frac{4!}{2!\cdot 2!}\cdot 5^3$ ways to have four out of seven people with two birthdays on Monday and two on Tuesday.

$P(\text{mmtt}) = \frac{^7C_4\cdot \frac{4!}{2!\cdot 2!}\cdot 5^3}{7^7} = .03187$

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  • $\begingroup$ Can you explain the procedure of the numerator. how did you apply the multiplication rule? $\endgroup$
    – Rayri
    Jul 14 '18 at 22:04
  • $\begingroup$ This looks more reasonable, but maybe you can explain the different factors and why they are there to those who are new to combinatorics? $\endgroup$ Jul 14 '18 at 22:07
  • $\begingroup$ 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? $\endgroup$
    – Rayri
    Jul 14 '18 at 22:20
  • $\begingroup$ Yes, $\frac{4!}{2!\cdot 2!}$ 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. $\endgroup$
    – Phil H
    Jul 14 '18 at 22:23
  • $\begingroup$ @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. $\endgroup$
    – Phil H
    Jul 14 '18 at 22:29
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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 $${7\choose 2}\cdot {5\choose 2} \cdot 5^3= 7\cdot 3\cdot 5\cdot 2\cdot 5^3 $$ so the probability you seek is $$P = {30\cdot 5^3\over 7^6}$$

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  • $\begingroup$ However, I checked the solution and it was: 7!*5^3/24*7^7 $\endgroup$
    – Rayri
    Jul 14 '18 at 21:54
  • $\begingroup$ 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 $30$ $\endgroup$ Jul 14 '18 at 21:55
  • $\begingroup$ @Rayen then why did you click accept answer? $\endgroup$ Jul 14 '18 at 21:55
  • $\begingroup$ Because in fact there is nothing wrong with your answer. Thanks you a lot ❤️ $\endgroup$
    – Rayri
    Jul 14 '18 at 22:01

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