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Question Number 91230 by waseemyaqoobese@gmail.com last updated on 28/Apr/20

Σ_(j=o) ^m    ^a C_j   ^b C_(m−j)   =  ^(a+b) C_m   solve this problem

$$\underset{{j}={o}} {\overset{{m}} {\sum}}\:\:\overset{{a}} {\:}{C}_{{j}} \:\overset{{b}} {\:}{C}_{{m}−{j}} \:\:=\:\overset{{a}+{b}} {\:}{C}_{{m}} \\ $$$${solve}\:{this}\:{problem} \\ $$

Answered by ~blr237~ last updated on 28/Apr/20

Answered by mr W last updated on 28/Apr/20

we can get the coef. of x^m  in (1+x)^(a+b)   in two ways:  (1+x)^(a+b) =Σ_(m=0) ^(a+b) C_m ^(a+b) x^m   or  (1+x)^a (1+x)^b =(Σ_(j=0) ^a C_j ^a x^j )(Σ_(i=0) ^b C_i ^b x^i )  =Σ_(m=0) ^(a+b) (Σ_(j=0) ^m C_j ^a x^j C_(m−j) ^b x^(m−j) )  =Σ_(m=0) ^(a+b) (Σ_(j=0) ^m C_j ^a C_(m−j) ^b x^m )  ⇒C_m ^(a+b) =Σ_(j=0) ^m C_j ^a C_(m−j) ^b

$${we}\:{can}\:{get}\:{the}\:{coef}.\:{of}\:{x}^{{m}} \:{in}\:\left(\mathrm{1}+{x}\right)^{{a}+{b}} \\ $$$${in}\:{two}\:{ways}: \\ $$$$\left(\mathrm{1}+{x}\right)^{{a}+{b}} =\underset{{m}=\mathrm{0}} {\overset{{a}+{b}} {\sum}}{C}_{{m}} ^{{a}+{b}} {x}^{{m}} \\ $$$${or} \\ $$$$\left(\mathrm{1}+{x}\right)^{{a}} \left(\mathrm{1}+{x}\right)^{{b}} =\left(\underset{{j}=\mathrm{0}} {\overset{{a}} {\sum}}{C}_{{j}} ^{{a}} {x}^{{j}} \right)\left(\underset{{i}=\mathrm{0}} {\overset{{b}} {\sum}}{C}_{{i}} ^{{b}} {x}^{{i}} \right) \\ $$$$=\underset{{m}=\mathrm{0}} {\overset{{a}+{b}} {\sum}}\left(\underset{{j}=\mathrm{0}} {\overset{{m}} {\sum}}{C}_{{j}} ^{{a}} {x}^{{j}} {C}_{{m}−{j}} ^{{b}} {x}^{{m}−{j}} \right) \\ $$$$=\underset{{m}=\mathrm{0}} {\overset{{a}+{b}} {\sum}}\left(\underset{{j}=\mathrm{0}} {\overset{{m}} {\sum}}{C}_{{j}} ^{{a}} {C}_{{m}−{j}} ^{{b}} {x}^{{m}} \right) \\ $$$$\Rightarrow{C}_{{m}} ^{{a}+{b}} =\underset{{j}=\mathrm{0}} {\overset{{m}} {\sum}}{C}_{{j}} ^{{a}} {C}_{{m}−{j}} ^{{b}} \\ $$

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