How many combinations are possible between a note-event and a silence-event (i.e. a played note and a pause) over 16 consecutive beats? In other words, if one musical bar is divided into 16 parts (16th notes) of equal duration, filled with either a sound or a silence, how many combinations of the two are possible within the given structure of 16 events?   I hope this is clear enough...sorry, musician here, therefore almost completely unprepared to approach the question. I am not even sure if I am asking about combinations or permutations.   An example: (1= sound 0= pause) .  1000 0010 0011 0011  1111 0000 0111 1010
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How many distinct rhythms can a musical measure have?   Obviously the answer is not "$\infty$", so to answer this question we set a minimum rhythm  $\frac{1}{4}$. We will consider both notes and rests and not consider irregular groups like triplets and others. So we'll consider notes and rests as $\frac{1}{4 }$, $\frac{2}{4}$, $\frac{3}{4 }$ ... As regards musical measure we'll consider $\frac{L}{4}$ where $L$ is the length of measure and consider only $L&gt;0$. Then calculate the number of combinations of rhythms in a measure:     So the sequence found is: $1,2,5,13,34,...$ and this is a possible relation with Fibonacci's bisection: $$ F(0)=1 $$ $$ F(1)=2 $$ $$F(L)= 3 \cdot F(L-1) - F(L-2); \quad L&gt;1$$ How is it possible to proof that this recurrence relation is valid for each $L&gt;1$?  IMPORTANT EDIT: Sorry i have forgotten to write that we consider rhythms from the acoustic view point, so we consider for example that two rests of $\frac{1}{4}$ acoustically equal to a rest of $\frac{2}{4}$ and so we will consider only the rest of $\frac{2}{4}$.  Furthermore rhythms of arbitrary duration are allowed, for example we can have a note of $\frac{5}{4}$ with a invented symbol.
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The number of different ways to choose $k$ out of $n$ unique items:   Without repetition and without order-significance: $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ Without repetition and with order-significance: $\binom{n}{k}\times k!=\frac{n!}{(n-k)!}$ With repetition and with order-significance: $n^k$ With repetition and without order-significance: $?$   Thank you!
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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.
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How many different/unique $4$-letter arrangements are there of the letters in the word Mississauga?  I'm thinking we need permutations? I already tried finding each letter and still don't get it... any help?   So far I've broken it off to:-  $M=1$  $I=2$  $S=4$  $A=2$  $U=1$  $G=1$  However, I don't really know what to do next.
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Question:How many n-words from the alphabet {a,b,c,d} are such that a and b are never neighbors?   1.There are $4^n$ ways to arrange the four letters.  2.There are $(n-1)$$2^{n-1}$ ways of arranging a and b together.(is it so?)  3.Hence,the required number is $4^n$-$(n-1)$$2^{n-1}$.  Have I counted the right number of things?  I am looking for a very beautiful Combinatoric proof.     Edit:I realized I had incorrectly assumed that a and b occurred only once in the sequence.
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I gave my friend  as a brainteaser; while her attempted solution didn't work, it raised an interesting question.  I flip a fair coin repeatedly and record the results.  I stop as soon as the number of heads is equal to twice the number of tails (for example, I will stop after seeing HHT or THTHHH or TTTHHHHHH).  What's the probability that I never stop?  I've tried to just compute the answer directly, but the terms got ugly pretty quickly.  I'm hoping for a hint towards a slick solution, but I will keep trying to brute force an answer in the meantime.
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What is the shortest solution to the following problem?     What is the number of ways to order the 26 letters of alphabet so that   no two of the vowels a,e,i,o,u occur consecutively?   What I thought is to subtract permutations consisting of 2 vowels, 3 vowels, 4 vowels and 5 vowels occurring consecutively from all permutations 26!. But I could not even find an reasonable answer with that method. It seems to be too long and I always make mistakes.
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I have this question that states:  Using Inclusion/exclusion, find the number of derangements of each of the following strings.  A) aabcd  B) aabbcc  I understand inclusion/exclusion I just don't understand how it applies to this type of question.  I would appreciate the help and some clarification on how to go about solving these types of problems.
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Question: whats the probability that at least 3 out of a group of n people have the same birthday.  I am confused as to where to start. I know that it would be complementary probability, but even then I dont know exactly.  Would it be;  P(at least 3 have the same bday) = 1 - (|all 3 have different| + (2 have the same|)? Any directions and suggestions?
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Suppose we had a normal 52 count deck of cards. If two cards are randomly picked, what is the probability that both are Hearts given at least one Heart was chosen.  I think the answer would be $ \frac{12}{51}$ because there are twelve Hearts left after one is chosen with 51 total cards, but I might be approaching this problem the wrong way. Any idea on how I should think about it, or what I did wrong?
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${n\choose 2}$ counts the 2-combination $(a,b)$ where    $a\,\in\,\{\text{Set with n-k elements}\}\,b\,\in\,\{\text{Set with n-k elements}\}$ $a\,\in\,\{\text{Set with n-k elements}\}\,b\,\in\,\{\text{Set with k elements}\}$ Order trivial here? $a\,\in\,\{\text{Set with k elements}\}\,b\,\in\,\{\text{Set with k elements}\}$   ${k\choose 2}$ counts the 2-combination $(a,b)$ where    $a\,\in\,\{\text{Set with k elements}\}\,b\,\in\,\{\text{Set with k elements}\}$   ${n-k\choose 2}$ counts the 2-combination $(a,b)$ where    $a\,\in\,\{\text{Set with n-k elements}\}\,b\,\in\,\{\text{Set with n-k elements}\}$   $k(n-k)$ counts the 2-combination $(a,b)$ where    $a\,\in\,\{\text{Set with n-k elements}\}\,b\,\in\,\{\text{Set with k elements}\}$ First question, is that because the order is trivial, either $k(n-k)$ or $(n-k)k$ would suffice? Second question, under what circumstances will both $k(n-k)$ and $(n-k)k$ have to be counted?
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I was wondering how it is possible for overtones to co-exist at the same time. When one pulls a string, it starts to oscillate and forms a standing wave with frequency $$f_0 = f$$ Additionally, infinite overtones oscillate at integer multiples of the fundamental frequeny:$$f_n = (n + 1) * f$$ with n being the nth overtone. Since all overtones are essentially standing waves too and standing waves are formed due to reflection and interference, wouldn't the existence of multiple overtones disturb the wave?   And not only overtones but any disturbance like interfering waves from the environment, wouldn't they hinder the creating of standing waves? And lastly (from a physical standpoint) why has the fundamental frequency the greatest amplitude?
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How can you find count of the total combinations possible if you have the count of unique combinations? Suppose I need to form a sum of $3$ using $\{1,2\}$ Unique combn. $(1+1+1)$, $(1+2)$ hence '$2$' unique but total is '$3$' i.e, $(1+1+1)$, $(1+2)$, $(2+1)$ I know the elements used to make the combination Note: Number of unique combinations (ie. 2 here) is known. How can I find the total number of combination given the unique combinations?
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can anyone help me with this: We are considering a symmetric random walk that ends if level 3 is reached or level -1 is reached. Start=0  What is the expected number of walks? So I am looking for: $E[{\tau}]$ with $\tau$=the stopping time.
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Say, we have 4 red balls, 4 blue balls, 4 green balls, 4 yellow balls. How many ways are there to form a sequence of 10 balls such that every color of ball occurs at least twice?  My thought: r = red b = blue g = green y = yellow * = undetermined  Every arrangments should in this form: rrbbggyy**  Case 1: two * are same  $\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}10!}{4!2!2!2!}$  Case 2: two * are different  $\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}10!}{3!3!2!2!}$  Result = Case 1 + Case 2  Any suggestions would be appreciated!
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The question is: Find a recurrence relation for number of ternary strings of length n that contain two consecutive zeros.    I know for ternary strings with length one, there are 0. For a length of 2, there is just 1 (00), and for a length of 3, there are 5 (000,001,002,100,200).    I did a similar problem, finding a relation for the number of bit strings of length n with two consecutive zeros: $$a_n = a_{n-1} + a_{n-2} + 2^{n-2}$$    Since you can add "1" to the end of all the $a_{n-1}$ strings, "10" to all the $a_{n-2}$ strings, and "00" any string of size $n-2$.    For the ternary string problem, I'm pretty sure you would replace the $2^{n-2}$ with $3^{n-2}$, but confused about the other terms of the relation. My guess is that it would have the coefficient $2$ in front of the other terms, since you can add either $1$ or $2$ to the end of $a_{n-1}$ and either $01$ or $02$ at the end of $a_{n-1}$.    So I believe the answer for the relation is: $$a_n = 2a_{n-1} + 2a_{n-2} + 3^{n-2}$$    How does that look?
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If you toss a coin $n$ times, there are $2^n$ possible sequences of heads and tails. Let $E_n$ be the set of sequences which do NOT contain two consecutive heads and $e_n$, the number of sequences in $E_n$.  Thus $E_3$ $=$ $\{$ $TTT$, $TTH$, $THT$, $HTT$, $HTH$ $\}$ and $e_3$ $=$ $3$.  How many elements of $E_n$ have $T$ as the first toss?  Please can anyone help me out here?