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Question Number 107659 by bobhans last updated on 12/Aug/20

$$\oint \mathbb{B}\mathrm{obhans}\oint$$$$\mathrm{If}\mathrm{all}\mathrm{the}\mathrm{letters}\mathrm{of}\mathrm{the}\mathrm{word}\mathrm{MISSISSIPPI}\mathrm{are}\mathrm{written}\mathrm{down}$$$$\mathrm{at}\mathrm{random},\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{all}\mathrm{the}\mathrm{four}\mathrm{S}\mathrm{appear}\mathrm{consecutively}\mathrm{is}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$

Answered by bemath last updated on 12/Aug/20

$$\divideontimes \mathcal{B}e\mathcal{M}ath\divideontimes$$$$n\left(A\right)=⊡⊡⊡⊡-------=8\times \frac{7!}{4!.2!}$$$$⊡=letterofS$$$$n\left(S\right)=\frac{11!}{4!.4!.2!}$$$$p\left(A\right)=\frac{8.7!}{4!.2!}\times \frac{4!.4!.2!}{11!}=\frac{8\times 24\times 7!}{11.10.9.8.7!}$$$$=\frac{24}{11.10.9}=\frac{4}{11.15}=\frac{4}{165}.$$

Commented by bobhans last updated on 12/Aug/20

$$\mathrm{yes}..\mathrm{correct}$$

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