All Questions
48
questions
0
votes
3
answers
87
views
Evaluate using combinatorial argument or otherwise :$\sum_{i=0}^{n-1}\sum_{j=i+1}^{n}\left(j\binom{n}{i}+i\binom{n}{j}\right)$
Evaluate using combinatorial argument or otherwise $$\sum_{i=0}^{n-1}\sum_{j=i+1}^{n}\left(j\binom{n}{i}+i\binom{n}{j}\right)$$
My Attempt
By plugging in values of $i=0,1,2,3$ I could observe that ...
- 9,569
3
votes
4
answers
160
views
Proving $\sum_{i=0}^n (-1)^i\binom{n}{i}\binom{m+i}{m}=(-1)^n\binom{m}{m-n}$
I am trying to prove the following binomial identity:
$$\sum_{i=0}^n (-1)^i\binom{n}{i}\binom{m+i}{m}=(-1)^n\binom{m}{m-n}$$
My idea was to use the identity
$$\binom{m}{m-n}=\binom{m}{n}=\sum_{i=0}^n(-...
- 87
4
votes
3
answers
120
views
Show $\sum_{i=0}^n{i\frac{{n \choose i}i!n(2n-1-i)!}{(2n)!}}=\frac{n}{n+1}$
How can this identity be proved?
$$\sum_{i=0}^n{i\frac{{n \choose i}i!n(2n-1-i)!}{(2n)!}}=\frac{n}{n+1}$$
I encountered this summation in a probability problem, which I was able to solve using ...
user1253495
0
votes
0
answers
52
views
How can I prove that $ \sum_{k=1}^{n}\frac{k\cdot P(n,k)}{n^{k+1}} = 1$? [duplicate]
The answer is difficult to me, I cannot figure out how to compute it.
$\sum_{k=1}^{n}\frac{k\cdot P(n,k)}{n^{k+1}}=1$
If someone can explain some technique to do it, I'd appreciate it. I tried to ...
- 1
0
votes
1
answer
51
views
Simplifying Expressions involving Sigma
After reviweing the solutions to a question involving the Biomial Theorem, I arrived at a step, where i was unsure how it occured.
Specifically, i was confused about the logic of:
k=0 -> k=1
n-1 -&...
0
votes
1
answer
76
views
find the smallest possible value of m so that there are real numbers $b_j$ satisfying $f_9(n) = \sum_{j=1}^m b_j f_9(n-j)$ for $n>m$
For an integer n, let $f_9(n)$ denote the number of positive integers $d\leq 9$ dividing n. Suppose $m$ is a positive integer and $b_1,b_2,\cdots, b_m$ are real numbers so that $f_9(n) = \sum_{j=1}^m ...
- 2,031
0
votes
2
answers
56
views
Rewriting $\sum_{{i,j=0}\:\:i\ne j}^n \binom{n}{i}\:\:\binom{n}{j}$ [duplicate]
Solve $$\sum_{{i,j=0}\:\:i\ne j}^n \binom{n}{i}\:\:\binom{n}{j}$$ This was a contest math problem which I was not able to solve.
My work: I was very unsure about how to approach this question. In my ...
- 860
1
vote
3
answers
177
views
calculate :$2000 \choose 2000$+....+$2000 \choose 8 $+$2000 \choose 5$+$2000 \choose 2$
my attempt:
let's put $2000 \choose 2000$+...+$2000 \choose 1001 $+...+$2000 \choose 8 $+$2000 \choose 5$+$2000 \choose 2$=$A+B$
with $A$=$2000 \choose 2000$+...+$2000 \choose 1001 $
and
$B$=$2000 \...
user1069990
4
votes
3
answers
751
views
Probability you end dice rolling sequence with 1-2-3 and odd total number of rolls
Here's a question from the AIME competition:
Misha rolls a standard, fair six-sided die until she rolls 1-2-3 in that order on three consecutive rolls. The probability that she will roll the die an ...
- 1,415
2
votes
1
answer
401
views
Simplifying the alternating sum of n squares
This question is based on a curious problem from Donald Knuth's The Art of Computer Programming, exercise 7 to chapter 1.2.1. It's stated as the following:
Formulate and prove by induction a rule for ...
- 842
2
votes
0
answers
78
views
value of $\frac{\sum_{k=0}^r{n\choose k}{n-2k\choose r-k}}{\sum_{k=r}^n {n\choose k}{2k\choose 2r} {(\frac{3}{4})}^{(n-k)}({\frac{1}{2}})^{2k-2r}}$ .
The question requires us to find the value of $\frac{\sum_{k=0}^r{n\choose k}{n-2k\choose r-k}}{\sum_{k=r}^n
{n\choose k}{2k\choose 2r} {\left(\frac{3}{4}\right)}^{(n-k)}\left({\frac{1}{2}}\right)^{...
- 97
3
votes
2
answers
67
views
Is this combinatorics summation equality true?
Recently I came upon the following equality, for natural numbers $n,k$ such that $n\ge k\ge0$:
$$\binom nk=\sum_{m=0}^{\min(k,n-k)}\binom km\binom{n-k}m$$
First of all, is this equality even true? It ...
- 2,847
2
votes
0
answers
60
views
Finding a formula for a sum that involves binomial coefficients
Is there a formula for this sum:
$$ \sum_{j=0}^k {n \choose j} {n \choose k-j} (-2)^j \left(-\frac13 \right)^{k-j} ?$$
It reminds me to Vandermonde's identity; but as you can see there is a slight ...
- 897
1
vote
2
answers
75
views
Finding a nice solution to $\sum_{a,b,c\ge 1, a+b+c=9}\frac{9!}{a!b!c!}=18150$
I am trying to find a nice way to solve for
$$\sum_{a,b,c\ge 1, a+b+c=9}\frac{9!}{a!b!c!}=18150$$
I managed to solved it (and verified by computer) by doing manually (on paper) on $7$ cases and got a ...
- 13
1
vote
3
answers
335
views
Question involving the sum $\sum_{k=0}^n(-1)^k\binom nk^2$
I've been tasked to prove the following equation is true:
$$\sum_{k=0}^n(-1)^k\binom nk^2=\begin{cases}0&\text{if }n\ \text{is odd}\\\displaystyle(-1)^m\binom{2m}m&\text{if }n=2m,m\in\mathbb ...
- 2,847