nswer-to-25929-by-binome-formula-1-x-n-m-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-lt-

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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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nswer-to-25929-by-binome-formula-1-x-n-m-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-lt-

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