Combinations, Binomial Coefficients, and Pascal's Triangle

Explanations 5

• Here, you might have already encountered Pascal's triangle. This explanation mentions how if we consider an 'addition triangle', then the kkk-th term in the nnn-th row is equal …

  By Kada Williams, Alex Federspiel

• Note that (2nn)=(2n)!n!⋅n!=2nn⋅2n−1n⋅2n−2n−1⋅2n−3n−1⋅2n−4n−2⋅…⋅21⋅11\binom{2n}{n}=\frac{(2n)!}{n!\cdot n!}=\frac{2n}{n}\cdot \frac{2n-1}{n}\cdot \frac{2n-2}{n-1}\cdot \frac{2n-3…

  By Kada Williams

• Prove that if ppp is a prime number, then ppp divides (x+1)p−(xp+1).(x+1)^p-(x^p+1).(x+1)p−(xp+1). Deduce Fermat's Little Theorem.

  By Kada Williams

• a) Prove that kkk divides (nkn)\displaystyle\binom{nk}{n}(nnk​). b) Prove that nnn divides (nk)⋅gcd(n,k)\displaystyle\binom{n}{k}\cdot \text{gcd}(n,k)(kn​)⋅gcd(n,k). (Source: Kö…

  By Kada Williams

• Prove the following estimate: 2(2m−1m−1)=(2mm)>4m2m.2\binom{2m-1}{m-1}=\binom{2m}{m}>\frac{4^m}{2m}.2(m−12m−1​)=(m2m​)>2m4m​.

  By Kada Williams