8

No. The string $122222\cdots 2$ is never able to be "solved" (using only addition, subtraction, and equals signs) for any length $n$ as any positioning of addition, subtraction, and equals signs will always leave the expression to the far left as an odd number while any other expression in the equality chain is always going to be an even number.


7

Not an answer, but a quick table with the number of turns until the recurrence of the initial position on rectangular grids small enough to be testable by computer. The ant is initially facing up (that is, along a column); this matters for non-square boards that may have different periods depending on whether the ant starts by facing in the "long" or the "...


6

Suppose in one instance you chose $1 \mapsto 1, 2 \mapsto 2, 3 \mapsto 3$ and then $4 \mapsto 1$. Then in another instance you chose the following $4 \mapsto 1, 2 \mapsto 2, 3 \mapsto 3$ and then $1 \mapsto 1$. But they are the same function. So you are overcounting.


6

Let's lay out all of our $8$ $b$'s and consider all the positions an $a$ can take. We aim to count how many $a$'s we can fit in there: $$\cdot \quad b \quad \cdot \quad b \quad \cdot \quad b \quad \cdot \quad b \quad \cdot \quad b \quad \cdot \quad b \quad \cdot \quad b \quad \cdot \quad b \quad \cdot $$ Each $a$ can only go into one of those $\cdot$s. ...


6

The probability you are asking for is: probability that $0$ kings are missing $*$ probability to chose a king given that $0$ kings are missing plus probability that $1$ king is missing $*$ probability to chose a king given that $1$ king is missing plus probability that $2$ kings are missing $*$ probability to chose a king given that $2$ kings are ...


5

There are $\binom {52}5$ possible poker hands. There are $4\times 7=28$ cards $≤8$, hence there are $\binom {28}5$ possible poker hands that do not contain any card higher than $8$. Thus there are $$D= \binom {52}5-\binom {28}5=2,500,680$$ possible poker hands that contain at least one card $>8$. This will be our denominator. To get the numerator: ...


4

Let $x_n$ denote your sum of the reciprocals of all products of subsets of $\{ 2, \ldots, n \}$ with at least 2 elements. Then $$ x_n + \left( \frac 12 + \ldots + \frac 1n \right) $$ is the sum of the reciprocals of all products of non-empty subsets of $\{ 2, \ldots, n \}$, and $$ \frac 11 + 2 x_n + 2 \left( \frac 12 + \ldots + \frac 1n \right) $$ is the ...


4

I preassume in this answer that drawn balls are not replaced. In both cases (distinguishable balls or not) the answer can be formulated as:$$P(F\cap S)=P(F)P(S\mid F)=\frac35\frac24=\frac3{10}$$where $F$ denotes the event that the first ball drawn is white and $S$ denotes the event that the second ball drawn is white.


4

Mainly use "stars and bars": A sequence of 10 stars and 2 bars can be interpreted as three non-negative numbers that sum to 10. There are $10+2\choose 2$ such sequences. However, three of these sequences would require a digit 10 and have to be subtracted. So $${12\choose 2}-3=66-3=63. $$


4

There are $kl$ products in $M$ since there are $k$ ways to choose $i\in S$ and $l$ ways to choose $j \in T$. For each $\phi$, the products $x_{\phi(1)}x_{\phi(2)}, \ x_{\phi(2)}x_{\phi(3)},\dots x_{\phi(2l)}x_{\phi(2l+1)}$ are all in $M$, which makes a total of $2l$ products. Moreover, every product in $M$ will appear an equal number of times in $\sum f(\...


4

The Most Misleading Patterns in Mathematics | This is Why We Need Proofs This video doesn't match the description in the last paragraph of the question, but it does start with $1,2,4,8,16,31$ and the Borwein integrals, and it is about intuition not always being enough.


3

With Wilson’s theorem, you can make all the non-divisible terms by $p$ vanish (you group together $1,2,\ldots,p-1$, then $p+1, \ldots, 2p-1$, and so on). Hence, mod $p$, $$\frac{1}{p}\binom{p^2}{kp}=\frac{p \cdot 2p \cdot \ldots \cdot (p-1)p \cdot p}{p \cdot \ldots \cdot kp \cdot p \cdot \ldots \cdot (p-k)p}=\frac{1}{p}\binom{p}{k}=\frac{1}{k}\binom{p-1}{k-...


3

There are $\binom 42=6$ ways to choose positions for the two black balls. For each of these choices, there are 2 ways to place the other two balls. The total number of arrangements is $$N = 2\binom 42 =12$$


3

$${n+2\choose 5} = 12{n\choose 3}$$ So $${(n+2)!\over 5!\cdot (n-3)!}=12{n!\over 3!(n-3)!}$$ So $${(n+2)(n+1)n!\over 5\cdot 4\cdot 3! (n-3)!}=12{n!\over 3!(n-3)!}$$ So $${(n+2)(n+1)\over 20} =12$$ Can you finish now?


3

I have a proof for both claims, but it’s fairly technical and I’ll elaborate later. First, simplify the second sum into $\binom{m(j+1)+k}{2k+1}-\binom{mj+k}{2k+1}$. Then, regroup the terms per value of $a$ in $\binom{a}{2k+1}$, and use binomial magic to rewrite it as $$f_m(n,k)=\binom{m(n+1)+k}{2k+1}+\sum_{j=1}^n{\frac{j}{n+1}(-1)^{n-j-1}\binom{2n+2}{n-j+1}...


3

Here's a partial result: If you allow for negative labels, every (labeled) tree on $n$ vertices can be labeled to be a full change graph in exactly $2^{n-1}(n-1)!$ ways. Call a connected subset of a graph's vertices "biconnected" if its removal leaves the graph connected. First note that the partitions of a tree into two connected subgraphs are exactly the ...


3

I assume that you have four men, three women, each person is distinct, and we are wanting to count the number of ways that they can line up such that there are at least three men in a row (i.e. at least one man who has another man to both his left and his right). I'm sure there are other ways to see the solution, but here I'll showcase one of them. One of ...


3

The number of $13$-letter words containing CHARITY is $26^6\cdot 7.$ You can start with any $6$-letter word and insert CHARITY in any of seven places. Not sure how you get $26^7\cdot 7!.$ Similarly, the number of ways to have HORSES in a $13$-letter word is approximately $26^7\cdot 8$, not $26^7\cdot 8!.$ But $26^7\cdot 8$ counts twice the cases where a ...


3

We can use the method of committee-forming. Consider a group of $3n + 1$ people, such that we want to choose a committee of $n$ people from that group. Note $\sum_{k=0}^{n} \binom{3n-k}{2n} = \sum_{k=0}^{n} \binom{3n-k}{n-k} = \binom{3n}{n} + \binom{3n-1}{n-1} + ... + \binom{2n}{0}$. Select one person. We can either choose for him to not be in the group ...


3

I do not think the probability of a color switch at a pair of given consecutive draws starting at the $i^{th}$ draw is independent of $i$. No, it is identical for all $i$ in $1$ to $b+w-1$. The probability for a colour switch after draw $i$ (call this event $S_i=1$), is the probability that draw $i$ is one colour and draw $i+1$ is the other. $$\forall i{\...


3

To count the number of ways: call the number of $b$'s to left of the first $a$: $x_1 \ge 0$ the number of $b$'s inbetween the first and two: $x_2 \ge 1$. the number of $b$'s inbetween the second and third: $x_3 \ge 1$. the number of $b$'s inbetween the third and fourth: $x_4 \ge 1$. the number of $b$'s to the right of the fourth : $x_5 \ge 0$. We have the ...


3

Maybe E. Trizac - When random walkers help solve intriguing integrals?


3

In how many ways can $5$ boys and $5$ girls stand in a queue if all five girls stand consecutively in the queue? You are correct that there are $6!5!$ ways for all five girls to stand consecutively in the queue. Method 1: We treat the block of five girls as a single object. We then have six objects to arrange, the block of girls and the five boys. The ...


3

Suppose the number begins with $3$. Then it must end with a digit small enough so that the product (which would be the middle digit) is $<10$. That admits $1+[9/3]=4$ choices where $[x]$ is the greatest integer less than or equal to $x$ and the possibility of a zero units digit is included. Do this for all nine possible initial digits and you get $9+[...


2

There is a rather similar approach that includes the 4! you were thinking of: Let's say you've got 4 balls with 4 different colors. Then there are 4! ways to permute those. Back to your case: Every single permutation has its permutation-buddy with the black balls switched. you can't make out any difference between those two permutations. Hence they are the ...


2

Your title says $P_{n+1}$ but your question asks for $P_{n-1}$. Assuming you are going for non-intersecting paths only. Consider a path with $n-1$ vertices. Once you fix the first vertex and the last vertex of the path, the remaining $n-3$ vertices can be put in different orders to create a unique path with each arrangement. So the number of such paths is: ...


2

Use the binomial theorem and you will find: $-560x^{4}$. The binomial theorem is: $$\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$$


2

The answer is $${28!\over7!7!7!7!}.$$ Put the items in a line in $28!$ ways, and put the first $7$ items in the first box, the next $7$ in the second box, and so on. Since the order the items are placed into a box doesn't matter, we have to divide by $7!$ once for each box.


2

Start by picking the group of three men who sit together. This can be done in $\binom{4}{3}$ ways. Consider this group as one object. We then have five objects to arrange: the group of three, the remaining man, and the three women. The men in the group of three can be arranged in $3!$ ways. So in all, we have $\binom{4}{3}5!3!$ possible arrangements. ...


2

There are $b+w-1$ consecutive pairs. Number the draws chronologically and for $i=1,\dots,b+w-1$ let $X_i$ take value $1$ if pair $(i,i+1)$ produces a switch. Let $X_i$ take value $0$ otherwise. Then $X:=\sum_{i=1}^{b+w-1}X_i$ denotes the total number of switches and the $X_i$ have equal distribution. With linearity of expectations and symmetry we find:$$...


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