Skip to main content

All Questions

0 votes
3 answers
88 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 ...
3 votes
4 answers
163 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(-...
4 votes
3 answers
121 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 ...
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 ...
2 votes
2 answers
174 views

How to show that $ \sum_{i=0}^n\binom{n}{i}(-1)^i\frac{1}{m+i+1}=\frac{n!m!}{(n+m+1)!} $

How would one show that $$ \sum_{i=0}^n\binom{n}{i}(-1)^i\frac{1}{m+i+1}=\frac{n!m!}{(n+m+1)!} ? $$ Any hint would be appreciated. Note: I tried to recognize some known formula, but since I don't ...
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
77 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 ...
24 votes
5 answers
687 views

What is $\sum_{r=0}^n \frac{(-1)^r}{\binom{n}{r}}$?

Find a closed form expression for $$\sum_{r=0}^n \dfrac{(-1)^r}{\dbinom{n}{r}}$$ where $n$ is an even positive integer. I tried using binomial identities, but since the binomial coefficient is in the ...
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 ...
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 \...
4 votes
3 answers
760 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 ...
2 votes
1 answer
403 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 ...
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)^{...
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 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 ...

15 30 50 per page