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=S = {a1,a2,ama_{1}, a_{2}, \dots a_{m}},

Clearly, if we find this set, afterwards, there is only 11 way of distribution, since the order is fixed.

Say, amongst the set SS, the integer rr comes PrP_{r} times, then:

r=0aPr=m\displaystyle \sum_{r=0}^{a} P_{r} = m , and Pr0P_{r} \geq 0

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

_Below is an interesting problem: _

We want to create a Divisible Sequence of length HH from a number NN. In a Divisible Sequence, every term (except the starting number) is a divisor of the previous term. Examples of Divisible Sequences of length 33 starting with 1010 are:


10,10,510, 10, 5

10,2,210, 2, 2

10,10,110, 10, 1

10,1,110, 1, 1

For primes p1,p2,ptp_{1},p_{2}, \dots p_{t}, obtain an expression for the number of divisible sequences starting with p1q1p2q2ptqtp_{1}^{q_{1}} p_{2}^{q_{2}} \dots p_{t}^{q_{t}}, of length k+1k+1.

0qii0 \leq q_{i} \forall i

Note: You might leave this expression in a sum or a product form.

Note by Jatin Yadav
7 years, 6 months ago

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1 vote

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can u explain it more easily ?

Vicky singh - 7 years, 6 months ago

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