All Questions
88
questions
3
votes
5
answers
218
views
Coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$
Find the coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$
I tried splitting the terms inside the bracket into two parts $1+x+\dots+x^9$ and $x^{10}$, and then tried binomial theorem, but that ...
- 565
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 ...
- 9,599
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
$$\binom{m}{m-n}=\binom{m}{n}=\sum_{i=0}^n(-...
- 87
1
vote
1
answer
70
views
Absolute Value Condition in Generalized Binomial Theorem
What is the coefficient of $\frac{y^3}{x^8}$ in $(x+y)^{-5}$? when $|\frac{y}{x}|<1$ (EAMCET 2020)
I am using the generalized binomial theorem
$$(x+y)^r=\sum_{k=0}^{\infty}{r \choose k} x^{r-k}y^...
- 1,834
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 ...
user1253495
1
vote
0
answers
43
views
binomial expression with $x^r$
Solve the series $$\sum_{p\ge0}\sum_{r\ge0}(1+x^{n-p+1})\left(\binom{p}{r}^2x^r\right)$$
What I am having problem here is accommodating for that $x^r$ otherwise $\sum_{r\ge0} \binom{p}{r}^2 $ is quite ...
- 931
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
2
votes
0
answers
119
views
Sum of square of terms in binomial expansion
We have the binomial expansion as this
$$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}$$
Is there a formula for $\sum_{k=0}^n {n \choose k}^2x^{2k}y^{2(n-k)}$,...
- 713
3
votes
1
answer
110
views
Alternative proof for ${pn \choose n}{qn \choose n} \ge {pqn \choose n}$ inequality
Reading this question, I saw that for $p,q,n$ positive integers the following inequality holds:
$${pn \choose n}{qn \choose n} \ge {pqn \choose n}$$
The inequality is not tight.
A simple combinatorial ...
- 2,103
2
votes
3
answers
95
views
Why doesn't $\binom{n}{a-b}\binom{n-(a-b)}{b}=\binom{n}{a}$?
We know that the multiplication principle states that if there are $a$ ways of doing $A$ and $b$ ways of doing $B$ once $A$ has happened, then there are $ab$ ways of performing both actions.
Keeping ...
- 6,560
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
6
votes
0
answers
111
views
To find $n$ such that the expansion of $(1+x)^n$ has three consecutive coefficients $p,q,r$ that satisfy $p:q:r = 1:7:35$.
To find $n$ such that the expansion of $(1+x)^n$ has three consecutive coefficients $p,q,r$ that satisfy $$p:q:r = 1:7:35$$
My work:
Suppose the consecutive coefficients are $\binom{n}{k-1}, \binom{n}...
- 12.7k
4
votes
1
answer
72
views
Simplify $\sum_{k = 0}^n\sum_{k_1+\ldots+k_{m}=n-k} \binom{nm}{k_1,\ldots,k_m,n-k_1,\ldots,n-k_m}$
Let $n$ and $m$ be positive integers. I want to find a formula for the following expression:
$$\sum_{k = 0}^n\quad \sum_{k_1+...+k_m=n-k} \quad \binom{n-k}{k_1,\ldots,k_m}\binom{nm}{n-k_1,\ldots,n-k_m,...
- 1,093
0
votes
4
answers
177
views
Proof that ${n \choose k}={k-1 \choose k-1}+{k \choose k-1}+{k+1 \choose k-1}+\cdots+{n-1 \choose k-1}$ [duplicate]
$${n \choose k}={k-1 \choose k-1}+{k \choose k-1}+{k+1 \choose k-1}+\cdots+{n-1 \choose k-1}$$
my attempt:
in the first let's write the relation between ${n \choose k}$ and ${n \choose k-1}$ and for ...
user998997
5
votes
3
answers
160
views
Proving an identity relating to binomial coefficients
This question is from the book Discrete mathematics and its applications, by K. Rosen, 6th chapter and 27th question.
Show that if n is a positive
integers $2C(2n, n+1) + 2C(2n, n) =
C(2n+2,n+1)$
...
- 5,409