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

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?

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.
• 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. Jul 14 '18 at 22:19
• You are both amazing tutors. Thanks you a lot Jul 14 '18 at 22:23

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.

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

• Can you explain the procedure of the numerator. how did you apply the multiplication rule? Jul 14 '18 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? Jul 14 '18 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? Jul 14 '18 at 22:20
• 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. Jul 14 '18 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. Jul 14 '18 at 22:29

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 $$P = {30\cdot 5^3\over 7^6}$$

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