No it's just the A's that have to be adjacent. I can see that the answer 302.400 for the dutch word is correct regarding just the unique words but the question was how many possibilities without regarding A1A2A3-bertionl and A2A3A1-bertionl as the same word.
Coming from the dutch word "SINAASAPPELS" (oranges) I can understand that the answer should be 302.400 being 10!/(3!.2!) as the SSS and PP have to be counted as interchangable. But the question stated clearly permutaties and not combinations. So A1A2A3-bertionl is not the same as...
Hi,
I have this problem in which a word (lets say "ABERRATIONAL") is given. Now how many permutations (no unique words) can be made given that the 3 A's have to be next to each other?
My solution: regard the 3 A's as one block, so we have 12-3+1 = 10 letters. This gives 10! permutations...
Is this true?
All urns have the same amount of balls to start with: a+b
Every time a ball is drawn there is probability a/(a+b) for the first urn and a/(a+b+1) or (a+1)/(a+b+1) for every next urn.
Depending on which color the ball has a or b is increased by one.
So it doesn't seem to...
Hi Techmologist,
I think you are right! I tried to substitute 2^(K+1)-2 for N in the probabilityformula and bring it to its limit but didn't succeed. your approach is much more elegant and intuitive I might say.
I didn't think of the Poisson-distribution either. Rather stupid considering I...
@ techmologist,
Yes, the question is really bugging me. How should we try to solve this equation? It is a bit hard doing this by numbercrunching as the bigger powers of 2 cannot be referenced anymore by excel or programming languages.
Marc
And, moreover, the probability seems to converge towards 0.6323 as K heads increases (with N = 2^(K+1)-2 as average number of coinflips to obtain a string of K heads)
Hi Mathman,
Thank you for your answer and the compact notation in logic of what it all boils down to.
Yes I know of course. What I wanted to know is how these compared probabilites all give a result closely around 0.64. I find it hard to believe that this is coin-cidental...(forgive the...
Coin tossing ---extended version---
We learned in previous posts that to obtain a string of at least K Heads, one has to flip a coin 2^(K+1)-2 times (N) on average (Markov chains), e.g. K = 3 HEADS, N would be 14.
On the other hand: flipping a coin 14 times and calculating the probability...
Hi CRGreathouse,
Yes, you understood the problem correctly and I thank you very much for your solution. This matches my Monte Carlo simulation results. The only thing is, how do I get the formula in "binomial" perspective to this problem. I thought of 1-P(no 5 AND no 4), this translated with...
How to calculate this certain probability in dicerolls?
Hi,
I am writing a game in which AT LEAST a 5 and 4 must be rolled with 2,3,4,5,6 and 7 dice. To calculate this probability I got stuck with the binomial formula nCk*p^k*q^(n-k).
I am looking for the formula that gives the solution to...