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Question Number 119896 by benjo_mathlover last updated on 27/Oct/20

Commented by bobhans last updated on 28/Oct/20

$$\left(ii\right)5girlsmustbetogether\iff 5!\times 8!$$

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

$$\left(i\right)$$$$12!=479001600$$$$$$$$\left(ii\right)$$$$5!8!=4838400$$$$$$$$\left(iii\right)$$$$◻G◼G◼G◼G◼G◻$$$$◻=zeroormoreboys$$$$◼=oneormoreboys$$$${(1+x+x^2+...)}^2{(x+x^2+x^3+...)}^4$$$$=\frac{x^4}{{(1-x)}^6}=x^4\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_5^{k+5}{x}^k$$$$k=3,C_5^8$$$$\Rightarrow C_5^8\times 5!\times 7!=33868800$$$$$$$$\left(iv\right)$$$$◻AGGGB◻$$$$◻=zeroormorestudents$$$${(1+x+x^2+...)}^2=\frac{1}{{(1-x)}^2}=\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_1^{k+1}{x}^k$$$$k=7,C_1^8$$$$\Rightarrow C_1^8\times 2!\times C_3^5\times 3!\times 7!=4838400$$

Commented by Ar Brandon last updated on 28/Oct/20

Thanks for correction, Sir.

Answered by bobhans last updated on 28/Oct/20

$$\left(iii\right)no2girlsareadjacent$$$$let◼:girls,◻:boys$$$$◻◻◼◻◼◻◼◻◼◻◼◻◻\Rightarrow C{}_5^8\times 5!\times 7!$$$$$$

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