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

_________________

Deepakh