6
votes
How many ways can $7$ professors and $5$ students be seated at this long rectangular table so that no student sits across from another student?
The problem with your solution is that you're not accounting for the possibility that more than one pair of students sits across from one another, so you're double counting those combinations. You ...
5
votes
How many arrangements we can make from the word "SINGAPORE" where the letters $E$ and $I$ do not occur together?
You can take all the possibilities and subtract the cases where the letters $E$ and $I$ occur together.
Hence the desired result can be expressed as $9! - 2\times 8! = 282240$. Thus your answer is ...
3
votes
Accepted
Number of permutations with 1 in even cycles
The number of such permutations should be
$$\sum_{2\leq k\leq n \\\text{$k$ is even}}\binom{n-1}{k-1}(k-1)!(n-k)!=
\sum_{2\leq k\leq n \\\text{$k$ is even}}(n-1)!=\lfloor n/2\rfloor (n-1)!$$
where $\...
1
vote
Does every sufficiently long string contain consecutive permutations of another string?
We shall construct a counterexample for an alphabet of $2$ letters (say, $a$ and $b$) and sufficiently large $n$ (one can compute the explicit value from various inequality constraints we will impose ...
1
vote
Edge coloring of complete bipartite graph
For each vertex in $B$, find the color which appears most frequently on the edges ending at that vertex. For each $i$, let $B_i$ be the set of vertices in $B$ whose most frequent color is color $i$. ...
1
vote
Accepted
Edge coloring of complete bipartite graph
I think the idea is that, for every vertex $x$ in $B$, there is at least one colour $i$ such that $x$ is adjacent to at least $|A|/r$ vertices of colour $i$ (if $x$ is adjacent to fewer than $|A|/r$ ...
1
vote
FInd remainder of $2^{2^{2024}}$ divided by $10$
There are several ways to solve this, but if it were me, I'd separately compute the number modulo $2$ and modulo $5$, then combine them with the Chinese Remainder Theorem to get a congruence modulo $...
1
vote
Accepted
Evaluating the series $\sum_{i=0}^{\infty} \sum_{j=0}^i \sum_{h=0}^j \frac{a^h(2a)^{j-h}(3a)^{i-j}}{(i+3)!}$
It is just an iterated Cauchy product of the absolutely convergent series
$$\sum_{n=0}^{+\infty} \frac{a^n}{(n+1)!} = \frac{e^a-1}{a}.$$
by itself. We get
$$\sum_{n=0}^{+\infty} \Big(\sum_{p+q+r=n} \...
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