All Questions
96
questions
3
votes
1
answer
164
views
Combinatorial interpretation for $\binom{n}{3}- \lfloor \frac{n}{3} \rfloor$
P2, RMO 2003, India
For any natural number $n\gt7$, prove that $\binom{n}{7}-\lfloor \frac{n}{7} \rfloor$ is divisible by $7$.
My algebraic solution :
$$ \binom{n}{7} = \dfrac{n(n-1)(n-2)(n-3)(n-4)(n-...
- 10.6k
5
votes
2
answers
942
views
Relation between Pascal's Triangle and Euler's Number
My friends and myself were discussing Pascal's Triangle, specifically the following property of it.
First, consider the Pascal's Triangle -
$$1\\ 1\ 1\\ 1\ 2\ 1\\ 1\ 3\ 3\ 1\\ 1\ 4\ 6\ 4\ 1\\ 1\ 5\ ...
- 256
0
votes
1
answer
36
views
On powers of binomial coefficient
Consider $(1+n)^k$. Where both n, k are natural numbers.
We have binomial expansion $\sum_{i=0}^{k} \binom{k}{i}n^i $
Then we have for each i-th term, certain powers of n(at least $>=i$).
As i ...
0
votes
1
answer
45
views
A proof for sum including binomial [duplicate]
I was trying to prove some equation and reached this summation $\sum \limits_{j=0}^{m} {2j\choose{j}} {2m-2j \choose {m-j}} = 4^m $
I tried pascal identity and other known binomial identities but ...
user411780
0
votes
0
answers
48
views
Amount of Compositions of a Number n into k parts when the components are limited to the range [1;m] with m<n
The number of compositions of a number n into k parts is given by the binomial coefficent ${n-1 \choose k-1}$.
Is there a closed formula to this question, when the summands of the composition are ...
- 9
0
votes
1
answer
54
views
Identity on Theory of Numbers
Can anyone help me identify this identity? Or is there a known principle regarding this?
$k\binom{k}{k}-(k-1)\binom{k}{1}+(k-2)\binom{k}{2}-(k-3)\binom{k}{3}+\ldots +(-1)^{k-1}\binom{k}{k-1}$
Any ...
- 13
3
votes
1
answer
720
views
How many non-negative integer solutions exist for: $x+y+z=48$ where, $x<y<z$?
I want to find the number of non-negative integral solutions to the following:
$x+y+z=48$ where, $x<y<z$.
The answer is apparently 192 and the solution provided is $$\frac{\dbinom{50}{2}-\...
- 123
2
votes
1
answer
77
views
Digits patterns in power number
Definition
Given positive integers $a,b$, with $a>1$, let $D(a,b)$ be the sum of the base-$a$ digits of $b$.
In other words
Rearranging, we get: $b = r_{l} a^l + ... + r_2 a^2 + r_1 a^1 + r_0 a^...
- 2,697
4
votes
2
answers
6k
views
$n$ choose $k$ where $n$ is negative
I saw (in the book $A~ Walk~ Through~ Combinatorics$) that $\sum_{n \geq 0}{-3 \choose n} = \sum_{n \geq 0}{n+2 \choose 2}(-1)^n$, which confuses me. It seems that it can be derived directly from ...
- 43
6
votes
0
answers
221
views
What is the growth rate of the products of binomial coefficients?
Claim: Experimental data seems to suggest that
$$
{n \choose 1^a b}{n \choose 2^a b}{n \choose 3^a b}\cdots {n \choose m^a b}
\sim \exp\bigg(\frac{2n^{1 + \frac{1}{a}}}{ab+3b}\bigg)
$$
where $a$ and ...
- 20.8k
3
votes
2
answers
1k
views
Number of six digit numbers divisible by $3$ but none of the digits is $3$
Find number of six digit numbers divisible by $3$ but none of the digits is $3$
My try:
Let the six digits are $a,b,c,d,e,f$ such that
$$a+b+c+d+e+f=3p$$
where $1 \le p \le 18$
Now since $a \ge 1$...
- 10.2k
3
votes
2
answers
204
views
Prove that $x^n=\sum_{k=1}^{n}\sum_{j=1}^{k}(-1)^{k-j}\binom{k}{j}\binom{x}{k}j^n$
Prove that for every $x,n \in \mathbb{N}$ holds
$$x^n=\sum_{k=1}^{n}\sum_{j=1}^{k}(-1)^{k-j}\binom{k}{j}\binom{x}{k}j^n$$
This is so called MacMillan Double Binomial Sum, see Mathworld - Power, ...
- 558
2
votes
1
answer
219
views
Are there any power identities which don't belong to this list?
The problem of finding expansions of monomials, binomials etc. is classical and there is a lot of beautiful solutions have been found already, the most prominent examples are
Binomial Theorem, ...
- 558
3
votes
3
answers
200
views
A conjecture on the sum of binomial coefficients
I am looking for a proof, disproof or counter example of the following claim.
Let $C = \{k_1, k_2, \ldots, \}$ be a strictly increasing infinite sequence of positive integers which have a certain ...
- 20.8k
2
votes
2
answers
279
views
Concerning the identity in sums of Binomial coefficients
Let be the following identity
$$\sum_{k=1}^{n}\binom{k}{2}=\sum_{k=0}^{n-1}\binom{k+1}{2}=\sum_{k=1}^{n}k(n-k)=\sum_{k=0}^{n-1}k(n-k)=\frac16(n+1)(n-1)n$$
As we can see the partial sums of binomial ...
- 558