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# 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.

3 years, 2 months ago

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