What are the steps followed to get these probability answers? I understand most of the probability formulas but when doing questions I find it difficult to understand theoretically what the question is about and which formula to use for which. Can anybody please explain how to compute the answers? (are there any tricks how to know which formula to use for which?)


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*Four students go to a diner and order a hamburger, a cheeseburger, a fish sandwich, and a roast beef sandwich (one sandwich for each). When the waiter returns with the food, she forgets which student ordered which item and simply places a  sandwich before each  student. In how many ways can the waiter do this? (the answer is $24$)

*In how many ways can three men and two women line up for a group photo? In how many ways can they line up if a women is to be at each end? (I think the answer might be $5!$ and $3!$) 

*a club has $12$ members, in how many ways can the offices of president, vice president, secretary, and treasurer be filled if the president and the vice president must be different member? (the answer is $19008$)

*Colored flags arranged vertically on a flagpole indicate a signal (or massage). How many different signals are possible if $2$ red, $2$ green, and $3$ yellow flags are available and all the red and green flags and at least one yellow flags are used? (the answer is $330$)

*a poker hand consists of $5$ cards from a deck of $52$ playing cards. The hand is a “full house” if there are $3$ cards of one domination and $2$ cards of another. For example, three $10$’s and $2$ jacks form a full house. How many full house hands are possible? (the answer is $3744$)
Thank You.
 A: All five of these problems use the same basic principles. The most important one is that when you make a sequence of $n$ choices, and the $k$-th choice can be made in $w_k$ ways, the there are $w_1w_2\dots w_n$ different ways to make the whole sequence of choices. This explains (as explained in the first problem) why there are $n!$ different ways to line up $n$ objects. It’s also important to know without having to think too hard about it that any set of $n$ things has $\binom{n}k$ $k$-sized subsets.


*

*Imagine that the students are sitting in a row. When the waitress places a sandwich before each student, she’s simply arranging the sandwiches in a line, so the question is just asking how many different ways there are to arrange four things in a line. That’s $4!$: there are $4$ ways to choose which sandwich goes first, then $3$ ways to choose which of the $3$ remaining sandwiches hits the table next, then $2$ ways to pick the third sandwich, and only $1$ way (some choice!) to pick the last, for a total of $4\cdot3\cdot2\cdot1=4!$ ways.

*Yes, there are $5!$ ways to arrange the $5$ people in a line; the reasoning is as in the first problem. If the women are to be on the ends, there are $3!$ ways to arrange the $3$ men in the middle, but there are also $2$ ways to place the women: either woman can be at either end of the line. Thus, there are $2\cdot3!=12$ such lineups altogether.

*We use the same kind of reasoning yet again. There are $12$ possible choices for president. Once the president is chosen, there are $11$ possible choices for vice president, since the president and vice president cannot be the same person. But there is no such restriction on the secretary or the treasurer: each of them can be any of the $12$ people, so each can be chosen in $12$ ways. The grand total is therefore $12\cdot11\cdot12\cdot12=11\cdot12^3$.

*There are three cases, depending on whether one, two, or three yellow flags are used.  


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*Suppose that just one yellow flag is used, so that we’re arranging $5$ flags. There are $\binom52$ ways to choose which $2$ positions in the string of $5$ will contain the red flags. That leaves $3$ open positions, and there are $\binom32$ ways to pick $2$ of them for the $2$ green flags. The lone yellow flag will then fill the remaining slot $-$ no choice there. The total number of such $5$-flag signals is therefore $\binom52\binom32=10\cdot3=30$.  

*Suppose that two yellow flags are used. We reason in exactly the same way. There are $\binom62$ ways to choose the $2$ positions to be filled by red flags, and then there are $\binom42$ ways to choose which $2$ of the remaining $4$ slots will get the green flags; the yellow flags get the leftovers. The total number of such $6$-flag signals is therefore $\binom62\binom42=15\cdot6=90$.  

*I’ll leave it to you to finish off the problem by calculating the number of $7$-flag signals; just follow the same reasoning. When you’re done, add up the numbers from the three cases to get the final answer.


*There are $13$ denominations altogether, from ace through king, so there are $13$ ways to choose the denomination of the triplet. The pair has to be of a different denomination, so there are $12$ ways to pick its denomination. Thus, there are $13\cdot12$ possible types of full house. However, each type (e.g., three $10$’s and two jacks) can occur in several different ways. There are $4$ cards of each denomination, one for each of the four suits, so once its denomination is known, there are $\binom43$ ways to choose the three cards for the triplet. Similarly, there are $\binom42$ ways to choose two cards of the pair’s denomination. Thus, each type of full house comprises $\binom43\binom42$ different full houses of that type, and the grand total of possible full houses is $$13\cdot12\cdot\binom43\binom42=13\cdot12\cdot4\cdot6=3744\;.$$
A: *

*In this case we should use formula for number of permutations(http://en.wikipedia.org/wiki/Permutation): $A_n^k = \frac {n!}{n-k!}$. This formula uses when we have $k$ places and $n$ objects we should arrange in this places and the order of objects is important(i.e. situation when first person(for instance, John) will get hamburger and second person(for instance, Mike) will get cheeseburger differs from situation when Mike will get hamburger and John will get cheeseburger).

*It seems to me that in this task there is tne only difference between men and women but there isn't difference between persons who have the same sex. Then this case is the same as the case when we should line up three apples and two oranges. What do we have in this case? Firstly, we should notice that there is no difference if we swap two oranges or two apples. Then we can conclude that we need just two choose three places from all these five places, where we put men(or apples or something else). So, we must use formula for number of combinations (http://en.wikipedia.org/wiki/Combination): $C_n^k$ = $\binom {n}{k}$ = $\frac{n!}{k!(n-k)!}$. In this task we have $C_5^3$ and $C_4^3$ correspondingly.

*In this case we firstly allocate three posts - vice president, secretary and treasurer - between twelve persons. For this situation we must use formula for number of permutations with repetition: $n^k$ , where $k$ is number of places and $n$ is number of objects we need to put on this places. We use formula which takes into account repitition because one person can takes several posts. And in the end we need to multiply result by 11 because for each allocation of three posts we can assign somebody to post of president in 11 ways(i.e. 12 persons minus that person who was assigned to post of vice president). The result is $12^3*11$.

*Here we again use the formula for number of permutations from first item. So we put 8 flags on 8 positions. Next we notice that swapping of two flags with the same color doesn't change a message but we computed it twice. So if we have $k$ flags with same color, then different permutations of them doesn't change the message and consequently we must our result divide by number of permutations of these $k$ one-colour flags. As a result we have: $\frac {7!} {2!2!3!}$.

*It's also easy case. There are 13 denominations. Firstly we must find number of ways to allocate two of all 13 denominations into 2 groups: the first group for denomination which will have 3 cards chosen and the second one for denomination which will have 2 cards chosen. This number is equal $\frac {n!}{(n-k)!}=\frac {13!}{11!} = 13*12$. Also we must multiply this by number of ways to choose 3 cards from 4 card and number of ways to choose 2 cards from 4.

