Pascal's triangle has this famous hockey stick identity.
(n+k+1k)=k∑j=0(n+jj)
Wonder what would be the form for multinomial coefficients?
Answer
(a1+a2+⋯+ata1,a2,⋯,at)=t∑i=2i−1∑j=1ai∑k=1(a1+a2+⋯+ai−1+ka1,a2,⋯,aj−1,⋯,ai−1,k)
Pascal's triangle has this famous hockey stick identity.
(n+k+1k)=k∑j=0(n+jj)
Wonder what would be the form for multinomial coefficients?
Answer
(a1+a2+⋯+ata1,a2,⋯,at)=t∑i=2i−1∑j=1ai∑k=1(a1+a2+⋯+ai−1+ka1,a2,⋯,aj−1,⋯,ai−1,k)
I have injection f:A→B and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...
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