Question

The probability that in a group of $N$ people at least two will have the same birthday is

• A. $1-\frac{\left(365\right)!}{{\left(365\right)}^N(365-N)!}$
• B. $1+\frac{\left(365\right)!}{{\left(365\right)}^N(365-N)!}$
• C. $1-\frac{\left(365\right)!}{{\left(365\right)}^N(365+N)!}$
• D. None of these

Answer 1

The correct option is A $1-\frac{\left(365\right)!}{{\left(365\right)}^N(365-N)!}$

Each can have birthday in $365$ ways.
Son $N$ persons can have birthdays in ${\left(365\right)}^N$ ways.
Number of ways in which all have different birthdays ${=}^{365}{P}_N$
$P\left(A\right)=1-P\left(\overline{A}\right)=1-\frac{{}^{365}{P}_N}{{\left(365\right)}^N}=1-\frac{\left(365\right)!}{{\left(365\right)}^N(365-N)!}$