VeritasPrepKarishma wrote:
jamifahad wrote:
Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is
A. 341/12^6
B. 352/12^6
C. 352/12^5
D. 341/12^5
E. 371/12^6
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