# Combinatorics!

Hi, how do you hunt for the number of solutions of :

$$0 \leq a_{1} \leq a_{2} \leq a_{3} \dots \leq a_{m} \leq a$$.

Here is the solution:

Let $S =$ {$a_{1}, a_{2}, \dots a_{m}$},

Clearly, if we find this set, afterwards, there is only $1$ way of distribution, since the order is fixed.

Say, amongst the set $S$, the integer $r$ comes $P_{r}$ times, then:

$\displaystyle \sum_{r=0}^{a} P_{r} = m$, and $P_{r} \geq 0$

We know that the number of the solutions of this typical equation is ${m+a \choose a}$, hence , the set is chosen, and hence the required number of solutions of the original equation is also ${m+a \choose a }$.

_Below is an interesting problem: _

We want to create a Divisible Sequence of length $H$ from a number $N$. In a Divisible Sequence, every term (except the starting number) is a divisor of the previous term. Examples of Divisible Sequences of length $3$ starting with $10$ are:

$10,10,10$

$10, 10, 5$

$10, 2, 2$

$10, 10, 1$

$10, 1, 1$

For primes $p_{1},p_{2}, \dots p_{t}$, obtain an expression for the number of divisible sequences starting with $p_{1}^{q_{1}} p_{2}^{q_{2}} \dots p_{t}^{q_{t}}$, of length $k+1$.

$0 \leq q_{i} \forall i$

Note: You might leave this expression in a sum or a product form. 7 years, 6 months ago

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can u explain it more easily ?

- 7 years, 6 months ago