If you  choose 5 cards from 52 cards, shouldn't the answer be 52*51*50*49*48?  Why is the correct answer $\binom{52}{5}$? Any explanations will be appreciated!
120886
Can someone explain intuitively why this is 52 C 5?  What does it mean that the "order doesn't matter?" If I wanted to pick 5 cards from a 52 card deck, why wouldn't I just do 52!/47! since there are 52 spots for the first card, 51 for the second, 50 for the third, etc.?  Any help would be greatly appreciated!
53157
What is the probability that when you place 8 towers on a chess-board, none of them can beat the other.  Attempt: ${64 \choose 8}^{-1} \approx1$ in $4\ 400\ 000\ 000$  Correct answer: ${64 \choose 8}^{-1} \cdot 8! \approx 1$ in $9\ 000\ 000$.  I disagree with the $8!$. If there's combinations (binomial coefficient) in the denominator, why would there be permutations i.e. the order counts, in the numerator?
114348
Problem Statement:- The number of zeroes at the end of the integer   $$100!-101!+\ldots-109!+110!$$   I am having a bit of a trouble in thinking how do I proceed. A little push in the right direction would be appreciated.   And if you are posting a full solution do use the spoiler tag, as sometimes I cant stop myself from seeing the whole solution and lose the chance of thinking through it by myself with a push in the right direction from you guys.  Also, I dont know whether I am using the right tag feel free to correct it if its wrong.
16382
Four cards are face down on a table. You are told that two are red and two are black, and you need to guess which two are red and which two are black. You do this by pointing to the two cards you’re guessing are red (and then implicitly you’re guessing that the other two are black). Assume that all configurations are equally likely, and that you do not have psychic powers. Find the probability that exactly j of your guesses are correct, for j = 0, 1, 2, 3, 4. Hint: Some probabilities are 0. My professor worked out this example in class and I know the answers for j = 1 and j = 3 are zero, j = 0 is 1/6, j = 2 is 2/3, and j = 4 is 1/6 but I do not understand the process or concept behind the question. I do not understand where the numbers are coming from and why only the even js have a probability but not the odd js. The j guesses and pointing is confusing me. Can someone please help explain the question, I would appreciate it.
115103
So what I decided to do is to start a small case with only 3 people. There are three possible combinations that I could pair up the people, using the symbols A, B, and C to represent the persons involved: A&amp;B, A&amp;C, B&amp;C. The probability of each pair having the same birthday is $1\over365$ (the first guy gets any birthday, and the second guy only gets one birthday to choose from). What my intuition directed me to do next is to find the probability that any of the three events are true, or the union of the three events, which involved adding up the probabilities.   To put this generally, the equation for the probability SHOULD be:  ${n \choose 2} \div 365$  However, clearly this is not the case since $n \choose 2$ gets over 365 when $n = 28$. And with 28 people, there is obviously still a chance that they all have different birthdays (Person 1 has Jan. 1, Person 2 has Jan. 2. ... Person 28 has Jan. 28). Could anyone tell me what's wrong with my intuition? I don't want to know what's the right solution, I just want to know what's wrong with my solution .. it makes sense to my intuition, even though it's incorrect.
96204
I'm looking for some guidance in solving this. I have read through my text multiple times trying to find something similar to this and I'm coming up with nothing. I'm not really sure where to start on solving this problem.  The question simply states     Prove that ${{n}\choose{k}} = {{n - 2}\choose{k}} + 2{{n - 2}\choose{k - 1}} + {{n - 2}\choose{k - 2}}$.   I'm very new to binomial coefficients, and extremely lost.
60026
I realize that this question has been asked multiple times and I do not really want to know how to do it, I understand how to solve it, my issue is somehow I have used the distribution and I am not getting the right answer.  Method of Solving: I solved this in the first go by directly applying the distribution formula that is $$\binom {n+r-1}{r-1} $$ By using, $n = 10, r= 3$ we get, $$\binom{12}{2} = 66$$ however the answer given is $$\binom{12}{3} = 220$$ Next, I tried to solve by taking multiple cases that is, setting any variable, $x,y$ or $z$ to values from $0,1,2,3...10$, so for example if $x = 0$ then we get, $$y+z=10$$ now it is simple enough to distribute this even without the formula, we get $$\Bigl((0,10),(1,9),(2,8)(3,7),(4,6)\Bigl) \cdot 2,(5,5)$$ which gives us $11$ cases, that can be verified using the formula $$P_0 = \binom{11}{1}.$$ Now taking cases till $x=10$ we get, the final result as $$\sum _{n=0} ^{n=10} P_n = 11+ 10 + 9 + 8+ \ldots +1 = 66$$ Please tell me where I am going wrong or of it simply a case of the answer in the text being wrong, and if the second method is correct.
42162
Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.     What is the probability that he wears all three shirts after 6 days?   My try:   $$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$  First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.  Thanks in advance.
401
Could someone help me with how the following is calculated:  What is the probability of rolling a die seven times and getting at least one six?  My instinct told me it would be $1/6+\cdots + 1/6$ but this ends up being $7/6$.
29878
In the game of , it turns out that any seven cards contain a projective set. For fewer than 7 cards, how can we determine the probability that one or more sets exist (in terms of the number of cards)?  For those not familiar with Projective Set, here's a description from the wikipedia page linked above:   A Projective Set card has six binary attributes, or bits, generally represented by colored dots. For each color of dot, each card either has that dot or does not. There is one card for each possible combination of dots except the combination of no dots at all, making $2^6 - 1 = 63$ cards total.  Three cards are said to form a "set" if the total number of dots of each color is either 0 or 2. Similarly, four or more cards form a "set" if the number of dots of each color is an even number.  A card and itself could be said to form a two-card set, but as the cards in the deck are all distinct, this does not arise in actual gameplay.
116273
In the UK there are 90 bingo balls. A bingo card consists of 9 columns and 3 rows. A row contains exactly five numbers and four blanks. A column consists of one, two or three numbers and never three blanks. The columns are occupied by the number chosen from 1-9 (only nine), 10-19 (ten), 20-29 ... 70-79 and importantly 80-90 (eleven). Numbers in columns are ordered from top to bottom.  How many possible unnumbered card layouts are there?  How many possible numbered cards are there?  I have started to approach this problem by first choosing the blanks for the first two rows and then attempting to work out a compensating row. This was hard.  Next I begun to understand that there were a number of equivalence classes as columns could be reordered for the purposes producing different layouts.  Next I looked into the most convoluted card which had five numbers followed by four blanks in the first row and four blanks followed five number in the second. This satisfied the "no blank column" requirement. Now the final row could be arranged however it wanted, though there was one confusing overlap to deal with.   I considered another convoluted card which had as many columns of three as possible. It turns out it's possible have a maximum of three columns of three which also gives rise to the maximum of six columns of one. It also turns out you can have a maximum of six columns of two. I think the starting point might be to determine the classes of card initially by column combinations and then permute them.  Having gotten all muddled by the layouts, I turned to the numbers. With a single number, it was one of 9, 10 or 11 depending on the column and that was easy. For the two number case, let's say with a column of ten, the first ($i$) was chosen from 9 and the next from $10-i$. It suddenly occurred to me that it was just combinations, phew. How to fit that with the whole business of column choice lost me.  It seems possible that for each possible "set" of columns, there will be a way of determining the distribution of possible first columns and adding in an appropriate factor for the number combinations.  Frankly, it's all getting bigger than my coffee table.   Any simplifications from higher maths I'm missing?
44353
I was trying to answer      Find the probability of getting a full house from a $52$ card deck. That is, find the probability of picking a pair of cards with the same rank (face value), and a triple with equal rank (different from the rank of the pair of course).   My idea was this:  Take one rank. The number of ways you can get a pair from a rank is $C(4, 2)$.   Take another rank. The number of ways you can get a triple from that rank is $C(4, 3)$.   Therefore, the number of ways you can get a full house from just $2$ ranks is $C(4, 2) \cdot C(4, 3)$. But there are $13$ different ranks, and the number of ways you can select $2$ ranks out of $13$ is $C(13, 2)$. For each of these pairs of ranks, you can have $C(4, 2) \cdot C(4, 3)$ full houses. Therefore the number of possible full houses is $C(4, 2) \cdot C(4, 3) \cdot C(13, 2)$, and the probability of selecting a full house is  $\dfrac {C(4, 2) \cdot C(4, 3) \cdot C(13, 2)}{C(52, 5)} \approx 0.000720288$  However, this is way lower than the actual answer of $.00144$. Where did I go wrong here?
62763
I was wondering if someone could critique my argument here. The problem is to find the probability where exactly 2 people in a room full of 23 people share the same birthday.  My argument is that there are 23 choose 2 ways times $\displaystyle \frac{1}{365^{2}}$ for 2 people to share the same birthday. But, we also have to consider the case involving 21 people who don't share the same birthday. This is just 365 permute 21 times $\displaystyle \frac{1}{365^{21}}$. To summarize:  $$\binom{23}{2} \frac{1}{365^2} \frac{1}{365^{21}} P\binom{365}{21}$$
50570
You have a deck of 32 playing cards. Somebody draws one card after another and shows them to you. At any point of time you may bet that the next card is black. If it is indeed black you earn $10, otherwise nothing. If you don't do anything you earn nothing as well. Find the optimal strategy.   In other words you should find the point of time where the quota of remaining red cards in the deck is maximal.   Any hints how to to this?
26565
4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?  Here's how I tried to solve:  For the first draw, we have 52 cards, and we have to pick one suit. So, probability for this is $\frac{13}{52}$.  For the second draw, only 51 cards are left. The second suit has to be selected, so there are 13 cards from that suit. The probability is $\frac{13}{51}$.  Similarly, the third and fourth draw have probabilities $\frac{13}{50}$ and $\frac{13}{49}$ respectively.  Since the draws are independent, the total probability becomes $$\frac{13}{52} \times \frac{13}{51} \times \frac{13}{50} \times \frac{13}{49}$$  But my book says the answer is $\frac{{13\choose 1} \times {13 \choose 1} \times {13\choose1} \times {13\choose1}}{52 \choose 4}$.  My answer differs by a factor of $4!$. What did I do wrong?
66702
I was solving some combinatorics question and I came upon one I've been trying to solve for 2 days and I'm curios how to solve it.  It is known that in every set of 15 cars, at least 3 were manufactured from the same country (there can also be another 3 (or more) cars in the set which came from another country). prove that in every set of 100 cars, 15 were made manufactured in the same country.  Thanks to anyone who helps!
75040
Show that $(n-r) \binom{n+r-1}{r} \binom{n}{r} = n \binom{n+r-1}{2r} \binom{2r}{r}$.   In the LHS $\binom{n+r-1}{r}$ counts the number of ways of selecting $r$ objects from a set of size $n$ where order is not significant and repetitions are allowed. So you have $n$ people you form $r$ teams and select $r$ captains and select $(n-r)$ players.  The RHS divides up a team into 2 sets?