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If you denote by $B^n_m$ the number of ways to distribute n pens among m children, you can write down its generating series: $$\sum_{n=0}^\infty{B_m^nx^n}=(1+x+x^2+\ldots)^m=(1-x)^{-m}$$ Now by taking the coefficient at $x^n$ you find a formula for $B_m^n$: $$B_m^n=(-1)^n C_{-m}^n=C_{m+n-1}^n$$ The answer to the problem is $B^5_6 B^6_6 = C_{10}^5 ... 1 Your reasoning is correct, but the language in which you couch it is not. You write: My reasoning is that$a_1$and$a_{n-1}$themselves do not contain$011$, so the only way$011$could occur is if$a_1$starts with$0$and$a_{n-1}$starts with$11$. You’re speaking of$a_1$and$a_{n-1}$as if they were binary sequences, when in fact they are ... 0 Instead find the coefficient of$x^6 y^5$in$(1+x+x^2+x^3+...x^\infty)^6(1+y+y^2+y^3+....y^\infty)^5$I'll explain to you why taking the coefficient gives the answer for distribution problems. Consider the expression$(1+x+x^2+x^3+...x^\infty)^n$. You can write this expression as ... 2 It doesn’t. The coefficient of$x^6$in your polynomial is the number of ways to pick$6$of the$11$pens. For instance, the term$x^2\cdot x^4$(before you collect terms) corresponds to choosing$2$blue and$4$black pens. HINT: This problem is quite different. Let’s forget about the black pens for a moment and ask in how many ways$5$blue pens can be ... 0 Try splitting into two:$S_n = $number of sequences that start with$1$and$T_n = $the number those start with$0$. We have$a_n = S_n + T_n$. Now to get$a_{n+1}$from lower values: If we start with$1$, we can have$a_n$choices for the rest. This also means$T_{n+1} = a_n$. If we start with$01$, then we have$T_{n-1} = a_{n-2}$choices for the ... 1 For small enough versions of this problem, like the 4-bullet case that you're asking for here, you can get exact probabilities by writing down the equations on bullet speeds that give the outcome you're interested in, and then using Mathematica to integrate Boole[those equations] over$[0,1]^4$. To distinguish combinatorial types of outcomes, the useful ... 1 The$r$groups are labelled (though not explicitly stated) Now each object has$r$choices for group, and we just apply the multiplication principle to get$r\cdot r \cdot r.. = r^n$2 Your answer assumes that the lineman is distinguishable from the other positions, but that the other three positions are indistinguishable from each other. So, it solves the problem of choosing a lineman and three other players. But these three other players can choose their positions in$3!=6$ways. Their answer is naturally based off of permutations and ... 0 SOME HINTS: See that$\{1,2,3\}\cap\{2,4,6\}=\{2\}$so$P(A\cap B)=1/6$(doesnt matter what happen on yellow dice i.e.$P(A\cap B)=6/36$too). The sum for two rolls 5 means that you can take$1+4$,$2+3$,$3+2$or$4+1$then$P(C)=4/36$. Another example:$P(B\cap C)=2/36$because the only way to sum 5 with$a+b=5$where$a\in\{2,4,6\}$and ... 1 This certainly is far from a complete answer, but I hope it's helpful. I'll try to describe my thought process so it doesn't seem like this came out of nowhere. Experimentation I fixed$p=3$and initially tried to examine$a=0$and$a=1$by taking advantage of the corollary of Lucas' Theorem at the question you linked. Using black for "odd" and white for ... 1 None of the functions can be strictly increasing, so we count what I would rather call the monotonically non-decreasing functions. Such a function can be completely described once we know how many of$0,1,2,\dots,6$are mapped to$0$, how many are mapped to$1$, how many are mapped to$2$, and so on. For if$x_0$of them are mapped to$0$, it must be the ... 0 [I found this solution collaboratively with someone else offline.]$\def\nn{\mathbb{N}}\def\rr{\mathbb{R}}$Let$T(n) = ( \text{The theorem is true for any length-$n$ sequence from $\rr$} )$, for any$n \in \nn$. If$T(n)$is false for some$n \in \nn$: Let$m \in \nn$be the minimum such that$T(m)$is false [by well-ordering]. Let ... 0 Let's rephrase this question into a graph theory question. Basically you are asking if we have the directed complete graph,$K_n^*$, then there a decomposition into directed cycles all having length$n$. If you are not used to graph theory, the directed complete graph is a graph on$n$vertices where each vertex has a directed arc to all other vertices. ... 0 To explain André Nicolas's comment, for any group of$4$vertices, there is one interior intersection of the diagonals. Thus, for each of the$\binom{n}{4}$choice of$4$vertices, there is one intersection. Therefore, the maximum number of intersections (when there are no coincident intersections) is $$\binom{n}{4}$$ 0 Apply the bubblesort algorithm to the first sequence, using the second sequence as the definition of the sort order. In the bubblesort algorithm a correctly ordered pair is never swapped. 0 I discovered Hoeffding's inequality thanks to A.S.'s helpful comments. Using the notation on that page, the inequality I am trying to prove translates to:$P(H(n) < \frac{n}{2} - n^\epsilon) \lt \frac{1}{n^2}$Hoeffding's inequality directly gives the formula:$P(H(n) < \frac{n}{2} - n^\epsilon) \lt e^{-2 \cdot n^{2 \cdot \epsilon - 1}}$and since ... 0 You dont need partitions of integers, just pigeon-hole principle. If everything is indistinguishable there is only one way to place it, think that the empty box doesnt exist: n objects in n-1 boxes with no empty box (i.e. at least one object in every box). If you put one object in n-1 boxes you only have one free object more to put over one of these n-1 ... 0 If both objects and boxes are distinguishable: There are$n$ways to select a box that'll be empty. Since rest of the boxes will have at least 1 object each, therefore there will one box out of the$n-1$that will have two objects in it. There are$n-1$ways to choose this box. Further, we have${n\choose2}$ways to put two objects in the selected box and ... 0 Take all possibilities of arranging 7 defective packages amongst 20. The ones with fourth package on position 12 can be calculated by placing 3 packages among the first 11 and 3 among the last 8:${{11}\choose{3}} * {{8}\choose{3}} / {{20}\choose{7}}$(11 choose 3) * (8 choose 3)/(20 choose 7) = 0.11919504644 4 Consider$N=(1+\sqrt 3)^5+(1-\sqrt 3)^5$. This is an integer. Since$-1\lt 1-\sqrt 3\lt 0, we have $$0\lt -(1-\sqrt 3)^5\lt 1.$$ So, we have \begin{align}\lfloor (1+\sqrt 3)^5\rfloor&=\lfloor N-(1-\sqrt 3)^5\rfloor\\&=N\\&=(1+\sqrt 3)^5+(1-\sqrt 3)^5\\&=2\left((\sqrt 3)^0\binom 50+(\sqrt 3)^2\binom 52+(\sqrt 3)^4\binom ... 1 The distribution is not negative binomial, for in the negative binomial the trials are independent. Here we are not replacing after inspecting. The resulting distribution is sometimes called negative hypergeometric, but the term is little used. We can use an analysis close in spirit to the one that leads to the negative binomial "formula." We get the 4-th ... 2 For i=1,2 let W_i denotes the event that the marble taken form box i is white. For i=1,2 let B_i denotes the event that the marble taken form box i is black. Then the probability of success is:P((W_1\cap W_2)\cup(B_1\cap B_2))=P(W_1\cap W_2)+P(B_1\cap B_2)=P(W_1)P(W_2)+P(B_1)P(B_2)$$0 Hint: If O_2 is at the end (complement of event 2), does this affect the probability of event 1? 0 number of selecting children is {13 \choose 6} and 6 books can be arranged among themselves is 6! ways thus the answer is {13 \choose 6}.6!. 0 Hint: For i=1,\ldots,k, let A_i be the set of words that do not have the i^{th} letter. The universal set of all words has cardinality k^n. Use inclusion-exclusion to find the value you require, which is:$$\bigcap_{i=1}^{k}A_i^c.$$2 You should choose the six children that will receive the books, i.e., \binom{13}{6}, and then you should multiple it by the number of permutations of the six different books among the six children, i.e., 6!, thus \binom{13}{6}6! . 2 Exponential generating functions are useful for this sort of problem because we can find the answer by multiplying the EGF for each of the types of balls:$$\color{red}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}\color{green}{\frac12\left(e^x+e^{-x}\right)}\color{blue}{e^x}.$$This expands to$$\frac12\cdot\frac{x^2}{2!} + \frac12\cdot\frac{x^4}{4!} + ... 1 $$P(\text{having an accident})=$$ $$=P(\text{having an accident}\cap\text{being a bad guy})+P(\text{having an accident}\cap\text{being a good guy})=$$ $$=P(\text{having an accident}\mid\text{being a bad guy})P(\text{being a bad guy})+P(\text{having an accident}\mid\text{being a good guy})P(\text{being a good guy})=$$ ... 0 The information about who exactly is a physician and who is not isn't important. Let's take a simplified example. I show you a closed bag and say that inside there are 10 colored balls, numbered from one to ten. 7 of the balls are blue and 3 are red. If I use them to make all possible combinations of 4 balls, how many of them have exactly 3 red balls? ... 1 Except for the degeneraten=0$case, the number of chains is$4$times the$n$th Fubini number, A000670, also known as ordered Bell numbers. The linked Wikipedia articles lists various ways formulas for these numbers, the most elementary ones being $$a_n = \sum_{k=0}^n \sum_{j=0}^k (-1)^{k-j} \binom{k}{j}j^n \qquad\text{and}\qquad a_n \approx ... 2 In your counting method, you would count |||_ _ _ _ _ _ _ _ _ _ once but |_ _ _||_ _ _ _ _ _ _ three times, because you give 11 choices for each bar, so you would count the latter once for each combination of choices: 1_ _ _23_ _ _ _ _ _ _ (note that swapping 2 and 3 here corresponds to the same choices) 2_ _ _13_ _ _ _ _ _ _ 3_ _ _12_ _ _ _ _ _ _ ... 0 Repetition of numbers does not mean more than 1 bar at the same place (that means a zero), it means having bars equally spaced out. 3 If you think about it, the expression amounts to \binom{52}7\binom{45}7\binom{38}7\binom{31}7\binom{24}{24}, We could as well have left out the \binom{24}{24} part in the above expression for the residue as it will always evaluate to 1 But if you express it as \frac{52!}{7!7!7!7!24!}, the 24! in the denominator becomes mandatory, since by ... 2 Imagine drawing 52 empty boxes on a large sheet of paper. The first 7 will be the first player's cards. The next 7 for the next player, etc. And the final 24 slots are the leftovers. Now shuffle the cards and fill all the slots. There are 52! ways to do that. But we don't care what order the first 7 are in. So divide by 7!. We don't care what order ... 1 Call one 22-size group A and the other B. Since, it matters which is which, then the number of ways to make the three groups are$$\binom{60}{16}\binom{44}{22}\binom{22}{22} = \binom{60}{16,22,22} = 3.147908\times 10^{26},$$which is the same as the original answer. In other words, I choose 16 to be in the small group. Then I am free to choose 22 from the ... 1 Let's assume you are talking about natural numbers greater than one and positive factors. Lets say that the prime decomposition is p_1^{a_1}\cdot p_2^{a_2}\cdot\ldots\cdot p_n^{a_n} where of course n\in\Bbb N, n\ge 1, and for all a_k, a_k\in\Bbb N, a_k\ge 1. All the divisors of this number can be expressed with the same primes and with ... 1 Suppose that we have the persons 1,2,3,4,5,6 sitting next to eachother, with 1 sitting next to 6. Suppose person 1 has six cookies and he wants them to pass to person 4. He needs three times two cookies to pass three cookies to to person 2. So now person 6 doesn't have any cookies and person 2 now has three cookies. Person 2 can pass only ... 1 By the Inclusion-Exclusion principle, it is: 1-4*Pr(A is void in spades)+6*Pr(A,B are void)-4*Pr(A,B,C are void) A is void in spades with probability {39\choose13}/{52\choose13}. A,B are void in spades with probability {39\choose 26}/{52\choose26}. Step 1: 1-Pr(A is void)-Pr(B is void)-Pr(C is void)-Pr(D is void). Step 2: We double-removed the ... 0 a. 5!*5! If the kind of hat is wrong, we just get permutations. b. D5 * D5 (Where D5 is the number of derangements on 5 objects = [5!/e]) This derangement number is determined with the inclusion/exclusion method. c. 10*9*8*7*6. Without the correct hat/person limitations, the attendant can give the five fedoras to 10, 9, 8, 7, 6 persons respectively ... 0 The total number of permutations is 8!=40320, and this repeats each unique sequence 3!3!2!=72 times. Therefore there are \frac{40320}{72}=560 different sequences, and so RABARBAR appears with probability 1/560\approx0.0017857. 0 I would prefer the solution as 3!3!2!/8! Out of all 8! permutations, you can have any permutation of 3 A's, 3 R's and 2 B's. 3 There are 3 As and 8 possible spots for them, so the probability that they will be in the correct spots is \frac{1}{8\choose3}. Given that this has happened, there are 5 spots for the two B's so the probability that they are in the correct spot is \frac{1}{5\choose2}. Given that both of these have happened, the Rs must be in the correct spots, ... 0 The total ways are \frac{8!}{3!.3!.2!} so the probability of RABARBAR is \frac{3!.3!.2!}{8!} 1 a) We are choosing numbers one at a time. Whatever collection of numbers we chose, by symmetry the probability the first chosen number was biggest is \frac{1}{4}. We can also make a calculation based on your sample space \Omega. As you pointed out, this sample space has (10)(9)(8)(7) outcomes. We now count the favourables. There are \binom{10}{4} ... 3 There are only 3 basic patterns: 4-3-1,\; 4-2-2,\; and 3-3-2 Ways of distributing distinct objects to distinct boxes in these patterns is given by the multinomial coefficients,\dbinom{8}{4,3,1},\dbinom{8}{4,2,2},\dbinom{8}{3,3,2} However, since the persons are distinct, we also need to consider the permutations of the patterns which would be ... 0 Consider a combinatorial argument. Let \Lambda = \{a, b, c\} be our alphabet. We select the i positions in an n letter word for \{a, b\} characters in \binom{n}{i} ways and multiply by 2^{i} as we have a word (is each selected slot a a or is it a b?). The remaining n-i slots are all c's. Now we add up over all possible values of i: ... 3 we have$$\left(1 + x\right)^n = 1 + {n \choose 1} x + {n \choose 2} x^2+ \cdots + x^n \tag 1$$now putting x = 2 in (1) gives you the result. 4 In the binomial theorem,$$(x+y)^n={n \choose 0}x^{n-0}y^0+{n \choose 1}x^{n-1}y^1+{n \choose 2}x^{n-2}y^2+\cdots+{n \choose n}x^{n-n}y^n=\sum_{i=1}^{n} {n \choose i} x^{n-i}y^i$$Put$x=1$and$y=2$. 1 To be explicit: suppose we have already chosen$m$LI elements of$F_p^n$. These, then, form a basis to a subspace isomorphic to$F_p^m$. Now$|F_p^m| = p^m$. By the LI of the columns already chosen, none of the columns is the$0$-vector, so we obtain$p^m - 1\$ possible linear combinations we can make from our choices of columns, which we must exclude from ...