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Question Number 126534 by mohammad17 last updated on 21/Dec/20

Commented by mohammad17 last updated on 21/Dec/20

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Answered by mr W last updated on 21/Dec/20

1)  C_3 ^(18) ×C_4 ^(15) ×C_6 ^(11) =514 954 080    2)  −M−C−T−  3!×P_3 ^4 =144    3)  4^3 −1=63

$$\left.\mathrm{1}\right) \\ $$$${C}_{\mathrm{3}} ^{\mathrm{18}} ×{C}_{\mathrm{4}} ^{\mathrm{15}} ×{C}_{\mathrm{6}} ^{\mathrm{11}} =\mathrm{514}\:\mathrm{954}\:\mathrm{080} \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$$−{M}−{C}−{T}− \\ $$$$\mathrm{3}!×{P}_{\mathrm{3}} ^{\mathrm{4}} =\mathrm{144} \\ $$$$ \\ $$$$\left.\mathrm{3}\right) \\ $$$$\mathrm{4}^{\mathrm{3}} −\mathrm{1}=\mathrm{63} \\ $$

Answered by benjo_mathlover last updated on 22/Dec/20

(2) −_1^(st)  −^E −_2^(nd)  ^x −^A −_3^(rd)  ^x −^O −_4^(th)   = 3!×3!× (((4+1−1)),((       1)) )  =6×6×4=144

$$\left(\mathrm{2}\right)\:\underset{\mathrm{1}^{{st}} } {−}\overset{{E}} {−}\underset{\mathrm{2}^{{nd}} } {\overset{{x}} {−}}\overset{{A}} {−}\underset{\mathrm{3}^{{rd}} } {\overset{{x}} {−}}\overset{{O}} {−}\underset{\mathrm{4}^{{th}} } {−}\:=\:\mathrm{3}!×\mathrm{3}!×\begin{pmatrix}{\mathrm{4}+\mathrm{1}−\mathrm{1}}\\{\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$=\mathrm{6}×\mathrm{6}×\mathrm{4}=\mathrm{144} \\ $$

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