Probability of Non-Adjacent Zeros in 11-Digit Integer with 1 and 0 Digits

[Raghav Gupta]

So can we say you have understood and got the answer?

 

[haruspex]

That pattern is finding nth term for sequence
0, 1, 3, 8, 16, 27------

The last two terms are wrong, should be 19 and 43.

Haruspex dissapeared. He left me confused. I don't know how to proceed.

hey, I have to sleep sometime. (7:30am here now.)
Having got the recurrence relation, it is a trivial matter to work through from the smaller numbers:
f₀=1, f₁=2, so f₂=1+2=3, etc. up to f₁₀=144.
As to whether this method is too 'advanced', I would not have thought so. You don't need to know it is called a recurrence relation, and you don't need to solve the relation (i.e. find a general formula for f_n). It's more like applying induction. You need to spot that having filled in the first digit the problem for the remaining digits hasn't changed much.

The 10C1+8C3+... method is fine too. It involves rather more calculation though. If you spot the recurrence method, it's much easier.

 

[AdityaDev]

hey, I have to sleep sometime. (7:30am here now.).

we are in different timezones then. My 2:22 am is your 7:30 am!

 

[Raghav Gupta]

10C1+9C2+8C3+7C4+6C5

There should be 11_C0 term = 1 also considering no zeroes in number.

Having got the recurrence relation, it is a trivial matter to work through from the smaller numbers:
f₀=1, f₁=2, so f₂=1+2=3, etc. up to f₁₀=144.
As to whether this method is too 'advanced', I would not have thought so. You don't need to know it is called a recurrence relation, and you don't need to solve the relation (i.e. find a general formula for f_n). It's more like applying induction. You need to spot that having filled in the first digit the problem for the remaining digits hasn't changed much.

But how you got f₁₀ fast? It would take time to add that numbers starting from f₀ unless you use a software.
It's like fibonacci series where there are no first two terms 0, 1.
I think you may have applied Binet's formula? But it is also lengthy, considering the powers and not using calculator.

Haruspex left me. He made me confused that how to proceed.

How you know that Haruspex is he/she.
According to wikipedia-Haruspex

 

[AdityaDev]

How you know that Haruspex is he/she.
According to wikipedia-Haruspex

That does not matter. I got my doubt cleared and I am happy. Even if haruspex is a she.. what difference does it make?

 

[Raghav Gupta]

I was just joking.
But how we get f₁₀ fast? Do we have to add all terms starting from f₀

I got my doubt cleared and I am happy

I think you meant question and not doubt.

 

[haruspex]

But how you got f₁₀ fast? It would take time to add that numbers starting from f₀ unless you use a software.
It's like fibonacci series where there are no first two terms 0, 1.

f(0) = 1, f(1) = 2 easy
after that
1+2=3
2+3=5
3+5=8
5+8=13
8+13=21
13+21=34
21+34=55
34+55=89
55+89=144
Done. Isn't that simpler than working out all those combinatorial functions? But I suppose you used an app/spreadsheet for that.
It might not be obvious to you that f(0)=1, so start with f(1) and f(2) by hand instead.

One benefit of my approach is that it scales. You can easily deduce the asymptotic behaviour.
Putting the two approaches together yields an interesting result about the sum of a combinatorial series.

 

[Raghav Gupta]

Thanks, I learned other approach but what if there were more spaces instead of 10. Now should we use Binet's formula? In software we can make loops for calculating that nth term for a recurrence relation and here it's easy because there were only 10 spaces.
In my approach there is a sequence for finding nth term.

 

[haruspex]

Thanks, I learned other approach but what if there were more spaces instead of 10. Now should we use Binet's formula? In software we can make loops for calculating that nth term for a recurrence relation and here it's easy because there were only 10 spaces.
In my approach there is a sequence for finding nth term.

Even if there's no effort obtaining the combinatorial terms in your series, you still have to add n of them (or n/2, maybe). Using Fibonacci, you have to add n terms, but no combinatorials to calculate.
Ultimately, Binet may be fastest. You'd have to round to nearest integer, but I'm sure you could go a long way before that might go wrong.

 

[Raghav Gupta]

Yeah that's true, I agree with you. Ultimately it comes to the person in the first place how to originate approach from his mind.
I never have seen that approach by anyone in my school and it's a kinda gem for me.