# Permutation & Combination card sequence . .

I've been trying to do these 2 questions about Permutation & Combination which linked to card play.

Q1 says :

Q1 : A sequence of 5 cards is drawn from a standard deck of cards,
with replacement.

a) How many sequences will have at least one queen ?
b) How many sequences will have at least one king or one queen
( or both)?


for the first question i gave a try but i don't think is right :

a) the total cards = 52 and sequence of 5 cards and there 4 queen in the standard deck. There for 52^4 x 51 x 50 x 4 choices

b) total cards = 52 ^5 - 8x7x6x5x4x3x2x1 choices

Q2 : In the game of poker a player receives a subset of 5 cards,
called a poker hand, from the standard deck of 52 cards.
The order is which the cards are received is not important,
just the actual cards themselves.

a) How many poker hands contain 5 hearts ?
b) How many poker hands contain exactly 3 hearts ?
c) How many poker hands contain 2 clubs and 3 spades ?
d) How many poker hands contain exactly two suits ?


I didn't understand the second Question i found it kinda of complex.

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If in the question that is asked you encounter terms as 'at least' or 'at most' then always see that as an indication that the 'complement' is probably much easyer to handle. Instead of looking at (the number of) sequences with at least one queen look at (the number of) sequences with no queen at all. –  drhab Feb 14 '14 at 12:02

Q1a) $52^{5}$ sequences are possible (note the cards are replaced). $48^{5}$ sequences are possible without a queen. So $52^{5}-48^{5}$ sequences with at least one queen.

Q1b) $52^{5}$ sequences are possible. $44^{5}$ sequences are possible without a queen and without a king. So $52^{5}-44^{5}$ sequences with at least one queen or king.

Q2a) $5$ hearts are randomly chosen from $13$ hearts. This can be done on $\binom{13}{5}$ ways.

Q2b) $3$ hearts are randomly chosen from $13$ hearts and $2$ are randomly chosen from the remaining $39$ cards. This can be done on $\binom{13}{3}\binom{39}{2}$ ways.

Q2c) $2$ clubs from $13$ and $3$ spades from $13$. This can be done on $\binom{13}{2}\binom{13}{3}$ ways.

Q2d) Lets count this first for the suits clubs and spades. Then we find $\binom{13}{1}\binom{13}{4}+\binom{13}{2}\binom{13}{3}+\binom{13}{3}\binom{13}{2}+\binom{13}{4}\binom{13}{1}$ ways.

Here the first term stands for $1$ club and $4$ spades, the second for $2$ clubs and $3$ spades, et cetera.

There are $\binom{4}{2}=6$ choices for the pairs of suits so this amounts in: $6\times\left[\binom{13}{1}\binom{13}{4}+\binom{13}{2}\binom{13}{3}+\binom{13}{3}\binom{13}{2}+\binom{13}{4}\binom{13}{1}\right]$ ways.

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Hint:

In each case, note that it is easier to count the sequences which DON'T have the desired property:

For (a), you can count the number of sequences which do not contain at least one queen. This is the same as choosing a sequence of cards from the 48 non-queen cards in the deck! You can then use the fact that $$(\text{# with at least one queen})=(\text{total # of sequences})-(\text{# with no queens}).$$

Similarly, for (b), it is easier to count the number of sequences which contain neither a king nor a queen (as this is just a sequence chosen from the remaining 44 cards), then subtract.

For the second question, you deal with combinations instead of permutations. There are $\binom{52}{5}=\frac{52!}{5!(52-5)!}$ different hands in this case. I'll do (b) for you; maybe that will help you see how to do the others.

You wish to count the number of hands which contain exactly three hearts. We can choose which three hearts to use in $\binom{13}{3}$ ways, since there are 13 hearts in the deck. The remaining 2 cards must be chosen from the $39$ non-hearts cards; so, there are $\binom{39}{2}$ ways to choose those. Thus in total, there are $$\binom{13}{3}\binom{39}{2}=211,926$$ hands of this type.

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Note that although it is unusual for a card counting problem, Q1 clearly specifies that cards are drawn with replacement, so the total number of sequences is $52^5$. –  David Feb 14 '14 at 11:59
@David Whoops, right you are! I deleted my comment on that point. Thanks for the catch. –  Nick Peterson Feb 14 '14 at 12:33