For example,     $a + b + c = 4$   Solving this using stars and bars You have $4$ stars and $2$ bars:  $$ x | x | xx$$ For example.     Then what does $\binom{6}{2}$ mean? The number of arrangements of the two bars out of $6$ objects? If so, is this a general case?
121414
Let $a, b, c \in \mathbb{N}_0$. If $a+b+c=12$, how many solutions $(a,b,c)$ satisfy the equation?   Is the answer:      $729$
98444
I understand that the factorial gives the number of arrangements. For example: the factorial of zero i.e. an empty set ( doesn't occur) is 1.   As the empty set can be arranged only in 1 way - i.e. by filling nothing. Now, let's take an example: 5 distinct seats. How many ways 5 distinct seats can be arranged? - 5! ways i.e. 120.  So, basically, factorial gives us the arrangements. Now, the question is why do we need to know the factorial of a negative number?, let's say -5. How can we imagine that there are -5 seats, and we need to arrange it? Something, which doesn't exist shouldn't have an arrangement right? Can someone please throw some light on it?. I saw some research done on the formula -
60318
I'd like to put one symbol over another? There are some similar things in math, like $\binom{a}{b}$, but I don't want the parentheses, or $\frac{a}{b}$ without the division line.  The closest I've found is $\overset{a}{b}$ (from mathtools) but the a ends up smaller than the b.
96602
So the question has three parts and I'm stuck on the third, that's how it goes :  Claude wants to invite 5 friends to dinner in a pool of 11 friends.  a) How many ways can he do this?  My answer : $\binom{11}{5} = 462$  It's correct  b) 2 of these friends can't be separated from each other, how many ways?  My answer : $\binom{10}{4} = 210$  It's correct  c) 2 of them are mad at each other and can't be invited together, how many ways?  My answer : ''a) - b)'' = $\binom{11}{5} - \binom{10}{4} = 252$  It's false  The answer: 378    I can't seem to understand, plus there is no explanation in my textbook, it's straight the answer, so I'm completely lost. I've tried lots of different things but never got it right.   Thank you for helping me understand. Combinatorics are like hell to me.
9540
Fix integer $n\geq 1$. Consider the number $$1-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots$$where the sum continues as long as the lower number in the binomial is $\leq n$. Is there a way to simplify this sum?   The first few values are $1, 0, -2, -4, -4$.
52482
If I have a $w \times h$ matrix where each value is an integer  $0 \lt n \lt 20$,  how can I count the number of distinct configurations, where $2$ configurations are "distinct" if there is no way to reshuffle the rows and columns that would produce the same matrix?  Can this be counted with the stars and bars method?    For example, these are equal (we swapped a row, then a column):  0 0 0    2 0 4 0 2 4    0 0 0   but these are distinct (no way to swap rows or columns to produce the other):  0 0 0    2 0 0 0 2 4    0 4 0   It seems like there ought to be a way to count the rows or columns as "bins" and the values as balls.  I realize that in this case there are $18$ different colored balls, but even if the only values possible were $1$ and $0$, (ball or no ball) I can't see how to represent it as stars and bars.
55780
Shilov's Linear Algebra writes a matrix with two bars on each side rather than square brackets. I didn't find any mention of it with a quick Google search, and I can't see any other examples this double bar notation.  Does it mean something other than a matrix? Is it an older notation? Or is it something else?  Thank you
96642
What does the number 2 mean in this picture?    Thanks.   Sorry I'm not a math expert.  I need this to make a program in C to generate a trinomial triangle for a guy who asked it on StackOverflow (it is his homework). Gotta earn that reputation. You know how it is!
100338
I have a question with regards to combinatorics. I am supposed to show the following combinatorial identity: $\sum\limits_{r=0}^n\binom{n}{r}\binom{m+r}{n}=\sum\limits_{r=0}^n\binom{n}{r}\binom{m}{r}2^r$.   The algebraic way is to consider the coefficient of $t^m$ in the expansion of $(1+t)^n(1-t)^{-(n+1)}$. For $r\in\{0,1,\cdots,n\}$, we have the coefficient of $t^{n-r}$ in the expansion of $(1+t)^n$ to be equal to $\binom{n}{r}$, and the coefficient of $t^{m+r-n}$ in the expansion of $(1-t)^{-(n+1)}$ to be equal to $\binom{m+r}{n}$. By the addition and multiplication principles this will show that the coefficient of $t^m$ in the expansion of $(1+t)^n(1-t)^{-(n+1)}$ will be equal to $\sum\limits_{r=0}^n\binom{n}{r}\binom{m+r}{n}$.  On the other hand, we note that $(1+t)^n(1-t)^{-(n+1)}=(1+\frac{2t}{1-t})^n(1-t)^{-1}=\sum\limits_{r=0}^n\binom{n}{r}2^rt^r(1-t)^{-(r+1)}$. As the coefficient of $t^{m-r}$ is equal to $\binom{m}{r}$, this implies that the coefficient of $t^m$ in the expansion of $(1+t)^n(1-t)^{-(n+1)}$ is equal to $\sum\limits_{r=0}^n\binom{n}{r}\binom{m}{r}2^r$.  What I am thinking of, is to think of a way to prove this identity without the use of the generalized binomial series. I was trying to think of a combinatorial method to solve this problem. In particular, I was trying to prove the combinatorial identity as follows:  Firstly, let us be given a class of $n$ boys and a class of $m$ girls, where $n\leq m$. Choose a team $A$ of $r$ boys, and choose a team $B$ of $r$ girls, and finally choose any subset (may be empty) of boys from $A$ to clean the classroom, where $r\in\{0,1,\cdots,n\}$. For each value of $r$, we see that there are $\binom{n}{r}$ choices for $A$, $\binom{m}{r}$ choices for $B$, and subsequently there are $2^r$ ways to choose a subset from $A$. In this way we get the number of ways to do so to be equal to the expression on the RHS of the identity.  What I am struggling though, is to be able to count the number of desired ways in a way that will give me the expression on the LHS of the identity. I tried firstly to choose the boys to clean the classroom before choosing the sets $A$ and $B$, but I wound up with some messy expression that does not seem close to matching the expression on the LHS.  I would like to know if there is another way of looking at this method (or I could have been missing out on something), or if there is any other combinatorial method (or any other algebraic method that does not involve the use of generalized binomial series) that one could think of to show the desired identity.
104266
Please give me feedback on my answer to this question. Question: For all $ n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} $   is equal to $ \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1} $. Answer: False, since;  Let $a=1,  b=-1$  and $n=1$  ,$k=0$   , using Binomial expression.  $$(a+b)^n=  \sum   \binom{n}{k}  a^{k}b^{n-k}$$  Then $(1+-1)^{2(1)}=\binom{2}{0}1^{0}(-1)^{2-0}+\binom{2}{1}1^{1}(-1)^{2-1}+\binom{2}{2}1^{2}(-1)^{2-2}=\binom{2}{0}-\binom{2}{1}+\binom{2}{2}$  Thus :For all $n\geq1:  \binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n}   \ne \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1}$  .
86551
How many arrangements are there of MATHEMATICS with both T's before both A's or both A's before both M's or both M's before the E ?  Can someone also point to some online resource that has such practice questions ?  Thanks.  Edit:  I have some initial thought, please tell me if I am in the right direction :  For the case both T's before both A's :  We can picture the permutation as follows :  (7) T (7) T (7) A (7) A (7)  (Remaining letters: M H E M I C S - total 7)  Each of the brackets are empty spaces for now. We need to pick 7 out of the total 7 X 5 = 35 spaces; that will ensure both T's being before both A's.  So number of ways = C(35,7) X 7!/2! (because two M's are same - we need to divide by 2!)
2521
Let's have 2 numbers, N and K, where K divides N. The number of K-combinations from a given set S of N elements is a well known formula. Let's concatenate N/K groups (resulting in N elements) such as the resulting set is N. How many possibilities are there, i.e. what's the formula?  For instance:    N=4, K=2, C(4, 2) = 6 {1,2},{1,3},{1,4},{2,3},{2,4},{3,4}   The 3 possibilities are:     {1,2},{3,4}   {1,3},{2,4}   {1,4},{2,3}   I generated these combinations and I think the number goes like this:  (4, 2): 3 (6, 3): 10 (6, 2): 15 (10, 5): 126 (9, 3): 280 (10, 2): 945 (14, 7): 1716 (12, 4): 5775 (15, 5): 126126 (15, 3): 1401400   Apparently, the result always divides with (N-1).
40841
How many functions from set $\{1,2,3,\ldots,n\}$ to $\{A,B,C\}$ are surjections? $n \geq 3$  Attempt  I was hoping to count the number of surjections by treating $A,B,C$ like bins, and counting the number of ways to fill them with $1,2,3,\ldots,n$ such that no bins are empty. This is equivalent to putting two "separator" bars in between the $n$ numbers or $n-1 \choose 2$.  However, I think I am missing a step here: prior to putting in the separator bars, I should permutate the $n$ elements. So, multiply by $n!$.  The order of elements in the same bins do not matter. So I should be divided by the number of permutations of the elements in each bins. This is where I am stuck.  Additional Info  I made the question up myself. Feel free to modify the question into a more solvable form if it is too difficult/complicated.
41912
In     However I am unable to work this out myself. When I multiply things out I get $\frac{1}{nS_{xx}}(\sum_{j=1}^n x_j^2 -2n\bar{x}x_i+nx_i^2)$. For things to be true, the terms inside the parenthesis can be rearranged to be $S_{xx}+n(x_i-\bar{x})^2$. I tried rearranging the terms so $\sum_{j=1}^n x_j^2 -n\bar{x}x_i+nx_i^2-n\bar{x}x_i$, but I can't seem to get to the answer. There must be some form of $S_{xx}$ that I am not aware of that is buried in their somewhere. Any help would be appreciated.
95326
I'm trying to figure out a combinatorial proof for:  $$\displaystyle \sum_{k \mathop = 0}^n \binom {r+k} k = \binom {r + n + 1} n$$  I've tried the committee counting thing, but that didn't work.
36827
I am attempting to solve this counting problem through combinatorial argument. The following is the equation I am given:   $$\sum_{i=1}^n (i-1)(n-i) = \binom{n}{3}$$  I understand that the right-hand side of this equation represents a set of $n$-elements out of which we choose 3. For example I believe we can say suppose we have a group of $n$ people and we want to choose 3 out of $n$ to be in a committee. However I'm not sure how to express the left-hand side in words. If forming a committee is an appropriate way to tackle this problem then I know the left-hand side must utilize the addition and multiplication principles, but I don't know how to put it into words. Also my intuition tells me that in solving this we should first flip $$(n-i)(i-1)$$  Thanks!
53069
I need to proof this following equality :  $$\binom{m+n}{m}=\sum_{i=0}^m \left(\binom{m}{i}\binom{n}{i}\right)$$  This is what I did combinatoric proof:  Left : subset with $m$ members from $m+n$  Right : $\binom{m}{i} \rightarrow$ The number of subsets in set $m$ could also be written as $2^m$ AND $\binom{n}{i} \rightarrow$ the number of subsets with $m$ members from set $n$  In case $ n \leq m \implies \binom{n}{i}=2^n$  So this is what I came up with I still can't see why those expression are equal.  Any ideas? Thanks.