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.

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TopNewestcan u explain it more easily ? – Vicky Singh · 3 years, 2 months ago

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