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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 ... 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!} + ...

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

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: $$P((W_1\cap W_2)\cup(B_1\cap B_2))=P(W_1\cap W_2)+P(B_1\cap B_2)=P(W_1)P(W_2)+P(B_1)P(B_2)$$

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_ _ _ _ _ _ _ ...

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 ...

1

Except for the degenerate $n=0$ 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 $$a_n = \sum_{k=0}^n \sum_{j=0}^k (-1)^{k-j} \binom{k}{j}j^n \qquad\text{and}\qquad a_n \approx ... 1 The r groups are labelled (though not explicitly stated) Now each object has r choices for group, and we just apply the multiplication principle to get r\cdot r \cdot r.. = r^n 1 The distribution is not negative binomial, for in the negative binomial the trials are independent. Here we are not replacing after inspecting. The resulting distribution is sometimes called negative hypergeometric, but the term is little used. We can use an analysis close in spirit to the one that leads to the negative binomial "formula." We get the 4-th ... 1 None of the functions can be strictly increasing, so we count what I would rather call the monotonically non-decreasing functions. Such a function can be completely described once we know how many of 0,1,2,\dots,6 are mapped to 0, how many are mapped to 1, how many are mapped to 2, and so on. For if x_0 of them are mapped to 0, it must be the ... 1 This certainly is far from a complete answer, but I hope it's helpful. I'll try to describe my thought process so it doesn't seem like this came out of nowhere. Experimentation I fixed p=3 and initially tried to examine a=0 and a=1 by taking advantage of the corollary of Lucas' Theorem at the question you linked. Using black for "odd" and white for ... 1$$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})= ...

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