# Math Help - Combinatoire -Problem

1. ## Combinatoire -Problem

Find all naturels numbers then this numbers , in this order, is the terms of arithmetic sequence.
remark :

2. Quelques remarques...

1. "naturel" se traduit par "natural" et en anglais, pas de s aux adjectifs
2. ici, ce n'est pas "then", mais "such that"
3. "combinatoire" se traduit par "combinatorics", même si ici, ça fait plus allusion à "combination" (pour combinaison)
4. même en anglais, on conjugue les verbes il y a plusieurs termes, donc c'est "are". Même remarque pour "this numbers", le pluriel de this est "these"

Si tu ne sais pas comment traduire un mot mathématique, cherche la page wikipedia correspondante et prends la page en anglais (en sélectionnant la langue sur la gauche)

Pourquoi ne pas utiliser le latex du forum ?

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Note that $n\geq 6$, otherwise, it doesn't make sense.

If these terms belong to an arithmetic sequence, then let the constant progression be r. Note that depending on the value of n, r can be positive or negative.

Then $\exists k,k' \in\mathbb{N}^*$ such that :

$C_5^n-C_4^n=kr$

$C_6^n-C_5^n=k'r$

After several calculations, we arrive at :

$C_5^n-C_4^n=\frac{n!}{(n-4)!5!} \cdot (n-9)$

$C_6^n-C_5^n=\frac{n!}{(n-5)!6!}\cdot (n-11)$

n cannot be 9 or 11, because the difference cannot be 0.

$\frac{k'}{k}=\dots=\frac{(n-11)(n-4)}{6(n-9)} \quad (1)$

n cannot be 10, otherwise k'/k is negative, which is a contradiction to the assumption.
if n=12, then there is a problem, as $C_6^n\leq C_4^n \leq C_5^n$. Contradiction because the order counts. (intuitively, they have to be on the same "half" of n... look at Pascal's triangle to have an idea)

If these are consecutive terms of an arithmetic sequence, then by (1) the solutions are n=7 and n=14 (it's a matter of solving a quadratic equation)

If not, I think that all natural numbers are solutions, except 1,2,3,4,5,9,10,11,12

3. I think there is some confusion about the notation of combinatorics.

In my country $C_n^k=\frac{n!}{k!(n-k)!}, \ 0\leq k\leq n$

Using this formula we have:

$\frac{2\cdot 5!}{n!(5-n)!}=\frac{4!}{n!(4-n)!}+\frac{6!}{n!(6-n)!}\Rightarrow\frac{10}{5-n}=1+\frac{30}{(5-n)(6-n)}\Rightarrow$

$\Rightarrow n^2-n=0\Rightarrow n\in\{0,1\}$

4. Hello, dhiab!

Find all natural numbers such that: . $C^n_4,\:C^n_5,\:C^n_6$ form an arithmetic progression.
$d$ = common difference: . $a_2 - a_1 \;=\;a_3-a_2 \;=\;d$

We have: . $C^n_5 - C^n_4 \;=\;C^n_6 - C^n_5 \quad\Rightarrow\quad \frac{n!}{5!(n-5)!}- \frac{n!}{4!(n-4)!} \;=\;\frac{n!}{6!(n-6)!} - \frac{n!}{5!(n-5)!}$

Multiply by $\frac{6!(n-4)!}{n!}\!:\quad 6(n-4) - 30 \;=\;(n-4)(n-5) - 6(n-4)$

. . which simplifies to: . $n^2 - 21n + 98 \:=\:0 \quad\Rightarrow\quad (n-7)(n-14) \:=\:0$

. . and has roots: . $\boxed{n \:=\:7,\:14}$

The two sequences are:

. . $\begin{array}{ccccc}n = 7\!: & C^7_4,\:C^7_5,\:C^7_6 &=& 35,\:21,\:7 \\ \\[-3mm]
n = 14\!: & C^{14}_4,\:C^{14}_5,\:C^{14}_6 &=&1001,\:2002,\:3003 \end{array}$