Question

In how many ways can a poker hand have a suit missing (so no Spades, for instance)?

Answer 1

`C_(39,5)=575,757`

Explanation:

I'm going to assume that the poker hands in question are 5 cards.

A standard deck has 13 ordinal cards (Ace, 2-10, Jack, Queen, King) with each ordinal in 4 suits (Hearts, Diamonds, Clubs, Spades), for a total of 52 cards `(4xx13=52)`.

We'll work with the combinations general formula (because order doesn't matter), which is:

`C_(n,k)=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

If we were allowed to pick from the entire deck, we'd have:

`C_(52,5)=(52!)/((5)!(52-5)!)=(52!)/((5!)(47!))=>`

`(52xx51xxcancelcolor(orange)(50)^10xx49xxcancelcolor(red)48^2xxcancelcolor(brown)(47!))/(cancelcolor(orange)5xxcancelcolor(red)(4xx3xx2)xxcancelcolor(brown)(47!))=52xx51xx10xx49xx2=2,598,960`

But we're limiting ourselves to any card from 3 suits - or 39 cards and not the full 52, and so:

`C_(39,5)=(39!)/((5)!(39-5)!)=(39!)/((5!)(34!))=>`

`(39xx38xx37xx36xx35xx34!)/(5xx4xx3xx2xx34!)=(69,090,840)/120=575,757`