Five cards are dealt off a standard 52-card deck problem
Is this approach to the given problems correct?
Five cards are dealt off a standard 52-card deck. How many line-ups are there in which all 5 cards are of the same suit?
My approach:
There are four suits and thirteen card values, so we have four types of lists. Each type of list can produce $13*12*11*10*9$ lists. So there are $4*(13*12*11*10*9)=617,760$ possible line-ups in which all 5 cards are of the same suit.
Five cards are dealt off a standard 52-card deck. How many such line-ups are there in which exactly one of the 5 cards is a queen?
My approach:
We have five types of lists. For each type of list in which there is exactly one queen produces $51*50*49*48$ different lists. So there are $5*(51*50*49*48)=29,988,000$ possible line-ups in which exactly one of the 5 cards is a queen.
Thanks in advance.
$\endgroup$ 31 Answer
$\begingroup$Assuming line up means you only care about the first $5$ cards the solution to the first problem is correct. For the solution of the second you forgot to take away the queens and you also forgot to pick which of the four queens it is that you have.
If the order does not matter you need to divide by $5!$
Under are hidden solutions if the order does not matter:
Solution to the first problem: First select the suit in $4$ ways and then there are $\binom{13}{5}$ ways to choose $5$ from the $13$ of that suit. Final solution is $4\binom{13}{5}$
$\endgroup$ 7Solution to the second problem: First select which of the queens you want in $4$ ways. After that you must choose $4$ cards out of the $48$ non-queen cards, this can be done in $\binom{48}{4}$ ways. Hence the answer is $4\binom{48}{4}$
