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Question Number 103513 by bemath last updated on 15/Jul/20

coefficient of x^5  in expansion  (1+x^2 )^5 (1+x)^4  equal to   (a) 40      (b) 45     (c) 50    (d) 55  (e) 60

$${coefficient}\:{of}\:{x}^{\mathrm{5}} \:{in}\:{expansion} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{5}} \left(\mathrm{1}+{x}\right)^{\mathrm{4}} \:{equal}\:{to}\: \\ $$$$\left({a}\right)\:\mathrm{40}\:\:\:\:\:\:\left({b}\right)\:\mathrm{45}\:\:\:\:\:\left({c}\right)\:\mathrm{50}\:\:\:\:\left({d}\right)\:\mathrm{55} \\ $$$$\left({e}\right)\:\mathrm{60} \\ $$

Answered by OlafThorendsen last updated on 15/Jul/20

C_5 ^1 C_4 ^3 +C_5 ^2 C_4 ^1  = 5×4+((5!)/(3!2!))×4 = 60  ⇒ (e)

$$\mathrm{C}_{\mathrm{5}} ^{\mathrm{1}} \mathrm{C}_{\mathrm{4}} ^{\mathrm{3}} +\mathrm{C}_{\mathrm{5}} ^{\mathrm{2}} \mathrm{C}_{\mathrm{4}} ^{\mathrm{1}} \:=\:\mathrm{5}×\mathrm{4}+\frac{\mathrm{5}!}{\mathrm{3}!\mathrm{2}!}×\mathrm{4}\:=\:\mathrm{60} \\ $$$$\Rightarrow\:\left(\boldsymbol{{e}}\right) \\ $$$$ \\ $$

Commented by bemath last updated on 15/Jul/20

how do you got it ?

$${how}\:{do}\:{you}\:{got}\:{it}\:? \\ $$

Commented by OlafThorendsen last updated on 15/Jul/20

(1+x^2 )^5  = Σ_(k=0) ^5 C_5 ^k x^(2k)   (1+x)^4  = Σ_(k=0) ^4 C_4 ^k x^k   To obtain x^5  two solutions only :  x^2 ×x^3  or x^4 ×x^1   Then C_5 ^2 C_4 ^3 +C_5 ^4 C_4 ^1

$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{5}} \:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\mathrm{C}_{\mathrm{5}} ^{{k}} {x}^{\mathrm{2}{k}} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{4}} \:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{C}_{\mathrm{4}} ^{{k}} {x}^{{k}} \\ $$$$\mathrm{To}\:\mathrm{obtain}\:{x}^{\mathrm{5}} \:\mathrm{two}\:\mathrm{solutions}\:\mathrm{only}\:: \\ $$$${x}^{\mathrm{2}} ×{x}^{\mathrm{3}} \:\mathrm{or}\:{x}^{\mathrm{4}} ×{x}^{\mathrm{1}} \\ $$$$\mathrm{Then}\:\mathrm{C}_{\mathrm{5}} ^{\mathrm{2}} \mathrm{C}_{\mathrm{4}} ^{\mathrm{3}} +\mathrm{C}_{\mathrm{5}} ^{\mathrm{4}} \mathrm{C}_{\mathrm{4}} ^{\mathrm{1}} \\ $$

Answered by Rio Michael last updated on 15/Jul/20

(1 + x^2 )^5  = Σ_(r=0) ^5 ^5 C_r  x^(2r) ......(i)  (1 + x)^4  = Σ_(r=0) ^4 ^4 C_r  x^r ......(ii)  ⇒ (1 + x^2 )^5 (1 + x)^4  = Σ_(r=0) ^5 ^5 C_r x^(2r)  . Σ_(r=0) ^4 ^4 C_r x^r    when r = 1 for (i) and r = 3 for (ii) we get x^5   ⇒ possibility is:^5 C_1 .^4 C_3   also when r = 2 for (i) and r = 1 for (ii) we get x^5   ⇒ possibility is:^5 C_2 .^4 C_1   ⇒ all possibilities are:^5 C_1 .^4 C_3  +^5 C_2 .^4 C_1   an extended way of the solution above.

$$\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{\mathrm{5}} \:=\:\underset{{r}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\:^{\mathrm{5}} {C}_{{r}} \:{x}^{\mathrm{2}{r}} ......\left({i}\right) \\ $$$$\left(\mathrm{1}\:+\:{x}\right)^{\mathrm{4}} \:=\:\underset{{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\:^{\mathrm{4}} {C}_{{r}} \:{x}^{{r}} ......\left({ii}\right) \\ $$$$\Rightarrow\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{\mathrm{5}} \left(\mathrm{1}\:+\:{x}\right)^{\mathrm{4}} \:=\:\underset{{r}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\:^{\mathrm{5}} {C}_{{r}} {x}^{\mathrm{2}{r}} \:.\:\underset{{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\:^{\mathrm{4}} {C}_{{r}} {x}^{{r}} \: \\ $$$$\mathrm{when}\:{r}\:=\:\mathrm{1}\:\mathrm{for}\:\left({i}\right)\:\mathrm{and}\:{r}\:=\:\mathrm{3}\:\mathrm{for}\:\left({ii}\right)\:\mathrm{we}\:\mathrm{get}\:{x}^{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{possibility}\:\mathrm{is}:\:^{\mathrm{5}} {C}_{\mathrm{1}} .^{\mathrm{4}} {C}_{\mathrm{3}} \\ $$$$\mathrm{also}\:\mathrm{when}\:{r}\:=\:\mathrm{2}\:\mathrm{for}\:\left({i}\right)\:\mathrm{and}\:{r}\:=\:\mathrm{1}\:\mathrm{for}\:\left({ii}\right)\:\mathrm{we}\:\mathrm{get}\:{x}^{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{possibility}\:\mathrm{is}:\:^{\mathrm{5}} {C}_{\mathrm{2}} .^{\mathrm{4}} {C}_{\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{all}\:\mathrm{possibilities}\:\mathrm{are}:^{\mathrm{5}} {C}_{\mathrm{1}} .^{\mathrm{4}} {C}_{\mathrm{3}} \:+\:^{\mathrm{5}} {C}_{\mathrm{2}} .^{\mathrm{4}} {C}_{\mathrm{1}} \\ $$$$\mathrm{an}\:\mathrm{extended}\:\mathrm{way}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{above}. \\ $$

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