# Tag Info

## New answers tagged permutations

1 vote
Accepted

### In how many ways can we seat $3$ teachers and $8$ students in a row such that there are at least $2$ students between each teacher?

Let me first check whether the book answer is correct, then I shall check what is wrong with your answer $-TSSTSST-$ Place 2 students each between the 3 teachers, then there are 4 students left to ...
• 43.2k
Accepted

### Calculating number of permutations of an ordered set, only selecting in sequential order

There are $\binom 83=56$ ways to choose $3$ elements from a set of $8$ elements. For each distinct choice of $3$ elements, there is exactly one way to permute those elements to respect the ordering ...
• 24.4k

### Calculating number of permutations of an ordered set, only selecting in sequential order

Consider the following process: Take every permutation of 3 elements from the set. Sort each of those permutations. Group together all duplicate permutations. Can you see how the number of groups ...
• 25.9k
1 vote

### Number of 11 digit positive integers with non decreasing digits and finding the sum of digits of N

We can solve this question using beggar coin method also known as stars and bars method. Imagine you are given $3$ $9's$, $4$ $5's$ and $4$ $2's$. The number of non decreasing 11 digit numbers you can ...
• 155

### Show that every element of $A_n$ is a product of the cycles $(1\, 2\, 3), (1\, 2\, 4), \ldots, (1\, 2\, n)$.

Ugh, you CAN write any $(a~b~c)$ as a product of $(1~2~n)$, this is just a special case of a special case of $n\le 5$, right? We have several cases: $(a ~ b~ c)=(1~2~n)$, trivial. $(a ~ b~ c)=(2~1~n)$,...
• 2,591
1 vote
Accepted

### There are five students S1,S2,S3,S4 and S5 they are seated in maths

The word derangement means that nothing is in its original place. It sort of matches with non-mathematical usage of the word (that's how I remember it). If $S_1$ is in the right place guaranteed, we ...
• 218
Accepted

### Correspondence between elements of generalized symmetric group with different definitions

You can think of elements in $\mathbb{Z}_m\wr S_n$ as acting on the space $(R_m)^n$, where $R_m$ is the ring of $m$th roots of unity, by first doing a pointwise multiplication (via a primitive $m$th ...
• 4,824

### Mapping of points in 2d plane.

What you are asking is not reasonable. Visually, you cannot find a clear mapping between the points.
• 1,123

### working out dice probabilities with permutations/combinations

The result $\frac{5}{36}$ is correct. In case of doubt I would always resort back to looking at the individual draws to decompose the problem into manageable sub problems. Decomposition is a good ...
• 53

• 77.2k
Accepted

### Prove that $\frac{(n + 1)!}{((n + 1) - r)!} = r \sum_{i=r - 1}^{n} \frac{i!}{(i - (r - 1))!}$

If you divide both sides by $r!$, you find: $$\frac{(n + 1)!}{(n + 1 - r)! r!} = {n + 1 \choose r} = \sum_{i = r - 1}^{n} \frac{i!}{(i - (r - 1))! (r - 1)!} = \sum_{i = r - 1}^{n} {i \choose r - 1}$$ ...
• 7,727
### Prove that $\frac{(n + 1)!}{((n + 1) - r)!} = r \sum_{i=r - 1}^{n} \frac{i!}{(i - (r - 1))!}$
Induction on $n$ should work. For $n=r-1$ it is true and for the induction step note that $r!(\binom{n+2}{r}-\binom{n+1}{r})=r!\binom{n+1}{r-1}=r\frac{(n+1)!}{(n-r+2)!}$.