Diego

sup

Is that really the correct answer?

Then I saw the correct solution which uses stars and bars: There are \binom{15+4-1}{4-1} possible distributions for the passengers. Of all of them, only four correspond to the passengers getting off at a single stop(All at 1, All at 2...). Then the probability required would be 4/(\binom{18}{3})= 0.5% approx

In my first attempt, i said 'Well, each passenger, as a human being, chooses where to get off. So, for all to get off at stop 1, the probability would be (1/4)^15. Since there are 4 stops, the probability required would be (1/4)^14 which is approx. 0.0000003%

So, I had this problem of probability: In a bus there are 15 passenger and the bus makes 4 stops. Whats the probability that all 15 get off at the same stop?

hi

@LeakyNun I've got Dummit and Foot and it doesn't have comb

Does anyone have a recommendation on a book on groups that teaches applications to combinatorics? I already had a course on groups, but it was full abstract

does group theory have application on combinatorics?

@LeakyNun x^5-1 has 1 as solution.

?

@Albas How's that¡

want*

but again, this counts, for example, 1+1+1+1+96 as different from 96+1+1+1+1, and I don't wan't that.

whose formula is \binom{5+100-1}{100}

Then this is equivalent to put 100 balls in 5 numbered boxes.

And suppose I want non-negative integer solutions.

For example, suppose n=5.

I've just read the wiki and , as far as I see, it counts the tuples, and not the groups.

I really don't know the stars and bars method. I've always made combinatorial problems by "intuition" and knowing the classical formulas

Does it tell the number of distinct ones? I mean, I consider 1+2+97 the same as 97+1+2

I was wondering, is there any simple formula to know the number of distinct integer solutions to x_1+x_2+..+x_n=100?

hi

Hi

Thanks

This is the link: askubuntu.com/questions/773383/…

But I don't know how to "ping" it to you

Ok, done Thomas

would anybody here do me a favour? I wanted to comment a post to ask sthg but seems I don't have enough reputation... may someone comment by me?

Hi there

and never stops

it keeps giving me 6

?

anybody here¡