roof:Let B={integers set b+...+br+=n -1,where bi 2 0,≥k+1,i=2…,6+1≥0 f:A→B,(a1,…,ar)→(b1,…,br+1 By b1=a1-1≥0. b=a-a-1≥k+1,i=2,…,n 0r+1=n-a 0. Easy to check f is a bijection.And construct a function. g:B→C C={integers set c+...+c+1=n-1-(k+1)(r-1),where ci,....cr+1> 0,} andG=b,C=6-(k+1),i=2,·,r,Gr+1=br+1 Also,check g is a bijection.Hence, 4=1g=1C=（0-1-k+r)+r+)-=(（--)口 8.t.a nti omial Theorem) (1+2+…+xn)厂= +oa-alal…20.zn0 proof: (m1+x2+…+xn)y= the coefficient of n is equal to the number of vectors (veroecrtmsbproposition,weget th theorem proof: Let B ={integers set b1 + · · · + br+1 = n − 1, where b1 ≥ 0, bi ≥ k + 1, i = 2, · · · , r, br+1 ≥ 0.} f : A → B,(a1, · · · , ar) 7→ (b1, , · · · , br+1) By b1 = a1 − 1 ≥ 0. bi = ai − ai−1 ≥ k + 1, i = 2, · · · , r. br+1 = n − ar ≥ 0. Easy to check f is a bijection.And construct a function. g : B → C C ={integers set c1+· · ·+cr+1 = n−1−(k+1)(r−1), where c1, · · · , cr+1 ≥ 0,} and c1 = b1, ci = bi − (k + 1), i = 2, · · · , r, cr+1 = br+1. Also,check g is a bijection.Hence, |A| = |B| = |C| =  n − 1 − (k + 1)(r − 1) + (r + 1) − 1 r  =  n − k(r − 1) r  . Arrangements with Repetition Propostion8:X = {x1, · · · , xn}.B ={all vectors of length r over X s.t.xi occurs ai times.} Then,|B| = r! a1!a2!···an! . proof:just double counting. Corollary:(Polynomial Theorem) (x1 + x2 + · · · + xn) r = X a1+a2+···+an=r r! a1!a2! · · · an! x1 a1 · · · xn an . proof: (x1 + x2 + · · · + xn) r = X 1≤i1,i2,··· ,ir≤n xi1 xi2 · · · xir the coefficient of x1 a1 · · · xn an is equal to the number of vectors (i1, · · · , in) over [n],s.t.i occur ai times.by proposition8,we get this theorem. 7