Inclusion/Exclusion Principle Problem Help

[Guest]

Using the principle of inclusion/exclusion, how many ways are there to deal a six-card hand with at most one void in a suit?

Any help would be appreciated for this, I can't figure out where to start

 

[pka]

Using the principle of inclusion/exclusion, how many ways are there to deal a six-card hand with at most one void in a suit?

I know to how to answer the question. But I have no idea how the principle of nclusion/exclusion applies to this.
There are \(\displaystyle {{2+4-1} \choose 2}\) ways to deal six cards with no suit missing.
There are \(\displaystyle {4 \choose 1} {{3+3-1} \choose 3}\) ways to deal six cards with one suit missing.

 

[Guest]

It's helpful either way, so thank you very much.

 

[soroban]

Hello, b_as_a_constant!

Using the principle of inclusion/exclusion,
how many ways are there to deal a six-card hand with at most one void in a suit?

There are: \(\displaystyle \:{52\choose6} \:=\:\frac{52!}{6!46!} \:=\;\L20,358,520\) possible six-card hands.

The opposite of "at most one" is "two or more".


Void in two suits

There are \(\displaystyle \,{4\choose2}\) choices for the two suits.
Then 6 cards are chosen from the other two suits: \(\displaystyle \,{26\choose6}\) ways.
There are: \(\displaystyle \:{4\choose2}\,\cdot{26\choose6} \:=\:\frac{4!}{2!2!}\,\cdot\,\frac{26!}{6!20!} \:=\:\L1,381,380\) ways.

Note: This includes the hands which are void in three suits.


Therefore, there are: \(\displaystyle \,20,358,520\,-\,1,381,380\:=\:\L18,977,140\) hands
. . which are void in no suits or one suit.