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Question Number 25954 by abdo imad last updated on 16/Dec/17

nswer to 25929   by binome formula     (1+x)^(n+m)   =Σ_(k=0) ^(k=n+m)  C_(n+m) ^k  x^k     the coefficent of x^n  is   C_(n+m) ^n  and the coefficient of x^m   is   C_(n+m) ^m    but we have  for  p<n    C_n ^p    =   C_n ^(n−p)      >>>>>C_(n+m) ^n   =   C_(n+m) ^m .

$${nswer}\:{to}\:\mathrm{25929}\:\:\:{by}\:{binome}\:{formula}\:\:\: \\ $$ $$\left(\mathrm{1}+{x}\right)^{{n}+{m}} \:\:=\sum_{{k}=\mathrm{0}} ^{{k}={n}+{m}} \:{C}_{{n}+{m}} ^{{k}} \:{x}^{{k}} \:\:\:\:{the}\:{coefficent}\:{of}\:{x}^{{n}} \:{is}\: \\ $$ $${C}_{{n}+{m}} ^{{n}} \:{and}\:{the}\:{coefficient}\:{of}\:{x}^{{m}} \:\:{is}\:\:\:{C}_{{n}+{m}} ^{{m}} \:\:\:{but}\:{we}\:{have}\:\:{for} \\ $$ $${p}<{n}\:\:\:\:{C}_{{n}} ^{{p}} \:\:\:=\:\:\:{C}_{{n}} ^{{n}−{p}} \:\:\:\:\:>>>>>{C}_{{n}+{m}} ^{{n}} \:\:=\:\:\:{C}_{{n}+{m}} ^{{m}} . \\ $$

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