# Math Help - probability help

1. ## probability help

the question is something like this

a group of n girls is given names. there are N different names available. The girls are allotted names at random with 2 or more allowed to have same name. show the probability they will have n different names is (N-1)(N-2)....(N-n+1)/N^(n-1)

i think binomial distribution is involved init somewhere.

2. Hello, alex_2008!

A group of $n$ girls is given names.
There are $N$ different names available.
The girls are allotted names at random with 2 or more allowed to have same name.

Show the probability they will have $n$ different names is: . $\frac{(N-1)(N-2) \cdots (N-n+1)}{N^{n-1}}$
Let's "walk" through this . . .

. . . . $\begin{array}{ccccccc}\text{The 1st person can have any of the }N\text{ na{m}es:} & \qquad\quad Prob &=& \quad\dfrac{N}{N}\quad&=&\quad 1 \\ \\[-3mm] \end{array}$

$\begin{array}{ccccc}\text{The 2nd person can have any of the other }N-1\text{ na{m}es:} & \quad\;\; Prob &=& \dfrac{N-1}{N} \\ \\[-3mm] \end{array}$

$\begin{array}{cccccc}\text{The 3rd person can have any of the other }N-2\text{ na{m}es:} & \quad\;\; Prob &=& \dfrac{N-2}{N} \\ \\[-3mm] \end{array}$

$\begin{array}{ccccc} \vdots & & \vdots \\
\text{The }n^{th}\text{ person can have any of the other }N-n+1\text{ na{m}es:} & Prob &=& \dfrac{N-n+1}{N} \end{array}$

The probability is: . $1\cdot\underbrace{\frac{N-1}{N}\cdot\frac{N-2}{N} \hdots\frac{N-n+1}{N}}_{n-1\text{ fractions}}$

. . . . . . . . . . . . $= \;\frac{(N-1)(N-2)\cdots(N - n+1)}{N^{n-1}}$

3. thanks alot soroban