How many5-card hands can be dealt from a standard 52-card deck?different

I've highlighted the word different because I believe it's the key to the question, am I right in thinking the answer is

$\displaystyle 52*51*50*49*48 = 311,875,200$

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- Aug 29th 2012, 10:10 AMuperkurkBinominal cooefficient wording of a question
How many

5-card hands can be dealt from a standard 52-card deck?**different**

I've highlighted the word different because I believe it's the key to the question, am I right in thinking the answer is

$\displaystyle 52*51*50*49*48 = 311,875,200$ - Aug 29th 2012, 10:24 AMPlatoRe: Binominal cooefficient wording of a question
- Aug 29th 2012, 10:31 AMuperkurkRe: Binominal cooefficient wording of a question
Well I don't know that is just how the question is written but

23456

23465

23654

26543

65432

would all be classes as a different hand? I am not sure. - Aug 29th 2012, 10:49 AMPlatoRe: Binominal cooefficient wording of a question
- Aug 29th 2012, 10:54 AMHallsofIvyRe: Binominal cooefficient wording of a question
No, they wouldn't. Your answer, 52*51*50*49*48, which is the same as $\displaystyle \frac{52!}{47!}$ or $\displaystyle _{52}P_5$ which includes the same cards in different orders. To not count the same cards in different orders as different hands, you need to divide by 5!, the number of orders of the same 5 cards. That will give you $\displaystyle \frac{52*51*50*49*48}{5!}$ which is exactly $\displaystyle \frac{52!}{5! 47!}= \begin{pmatrix}52 \\ 5\end{pmatrix}$ as Plato said.

- Aug 29th 2012, 11:06 AMuperkurkRe: Binominal cooefficient wording of a question
ok thanks for explaining :)