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This is "stars-&-bars" - you're picking the points in the string where the 3 transitions are. So it's just $${10+3 \choose 3}=\frac{13!}{10!\:3!} = 286$$


This problem is equivalent to the number of ways to partition $10$ into four parts, in which the order of the partitioning matters. That is, $10 = 2 + 3 + 3 + 2$ is not the same as $10 = 3 + 2 + 2 + 3$. The first number represents the number of A's, the second number represents the number of C's, and so on. In order to then count the number of ways, think ...


There are six places GROOV can start. For each such place count the combinations of the other five letters, which should be $5!/2!$, since those five letters are distinct except for two O's. So $10!/4!$ overcounts by $6\cdot5!/2!$. So the final answer is $10!/4! - 6⋅5!/2!$.


When the coins are identical, all that matters is how many coins each person gets. So you are interested in the number of solutions to the equation $a+b+c+d+e=35$, where $a,b,c,d,e \geq 0$ are integers. The classical way of solving this equation is to imagine your 35 coins in a line, and four "separators" that separate the lot of each person. In total there ...


The theorem you are looking for is the Stars and Bars theorem, and more info can be found here: http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)


You are almost there. One problem with your way of thinking is that the same group will be counted multiple times. With $n=9$ and $k=4$ you have calculated $$ \frac{n!}{k!} = \frac{9!}{5!} $$ What you need is the binomial coefficient: $$ \binom{n}{k} = \frac{n!}{(n-k)!k!} = \binom{9}{5} = \frac{9!}{(9-5)!5!} $$ You get the extra $(n-k)! = (9-5)! = 4!$ ...


Firstly, the reason why exchanging rows and columns still maintains the condition we desire is because the condition does not care for where the letters are, only the number of occurrences of each letter. Swapping rows and columns clearly does not change that, so any row or column exchanges are okay. However, no matter how you switch the rows and the ...

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