Probability that the birthdays of 6 different persons will fall in exa

@fluke, please explain this :

000000,111111(All Birthdays in the same month)

Another slightly different way of approaching this (and a little more complicated than fluke's - which I think is a very appropriate solution. Nevertheless, it might be useful for people who have done permutation combination before).
Select 2 months and then no of people for each month. You select two months in 12C2 ways. Say you get J and F.
Then you say that J could have 1 person's bday, 2 people's bday, 3 people's bday etc

J - 1/2/3/4/5 (You select 1 or 2 or 3 or 4 or 5 people for J)
F - 5/4/3/2/1 (Corresponding to no of people selected for J, remaining people go in F)

The distribution can be done in \(6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 2^6 - 2 = 62\) ways (If you recognize it, great, else just calculate)

So probability \(= 12C2 * 62/ 12^6 = 341/12^5\)

My understanding of your solution goes this way.Let me know if I'm wrong.
1. Selecting any two months out of the 12 available can be done in 12C2 ways.
2. The 6 days can be distributed in the 2 available months in the following ways

M1 M2
0 6
1 5
2 4
3 3
4 2
5 1
6 0

which essentially is 6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6,
which is of the form
(x+a)^6 = x^6 + x^5 6C1 a^1 + x^4 6C2 a^2 + x^3 6C3 a^3 + . . + a^6.
Substituting 1 for 'x' and 'a' we arrive at (1+1)^6 = 1 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 1
2^6 = 2 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5
i.e, 2^6-2 = 6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 62
Here, we subtract 2 because we are not to take into account either cases where all 6 birthdays fall in M1 or M2

Thus the probability becomes 12C2 * 62/ 12^6 = 341/12^5

Exactly as the binomial dist you have written! That is basically weeding out all the cases which have "all" 6 birthdays in one month i.e. 6C6 and 6C0

Yes, that's correct. Many of us have used binomial theorem extensively and know that:
\(nC0 + nC1 + nC2 + ..... nCn = 2^n\) (an application of binomial as described by you)
Hence, it can be done this way.

On the other hand, you could obviously say that each of the six people have 2 options - either his bday falls in J or in F. So those are a total of \(2*2*2*2*2*2 = 2^6\) ways. Out of those, we need to remove 2 ways: All bdays fall in J and all bdays fall in F. Hence it can be done in \((2^6 - 2)\) ways.

Really mind boggling question, You can't solve it in one go unless and until you have solved such question before..
fluke & VeritasPrepKarishma very nice solution...
It seems easy once we read the solution but it's a real tough question..

Responding to a pm:

12C2 = 12*11/2

\(\frac{12C2 * 62}{12^6} = \frac{12*11*62}{2*12^6} = \frac{11*31}{12^5} = \frac{341}{12^5}\)

How did n(s) become 12*6

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