Question Number 124826 by bramlexs22 last updated on 06/Dec/20
$$\:{How}\:{many}\:{ways}\:{are}\:{there}\:{to}\:{arrange}\: \\ $$$${the}\:{letters}\:{of}\:{the}\:{word}\:'\:{VISITING}' \\ $$$${if}\:{no}\:{two}\:{I}'{s}\:{are}\:{adjacent}\:? \\ $$
Answered by mr W last updated on 06/Dec/20
$${Method}\:{I} \\ $$$${arrange}\:{at}\:{first}\:{the}\:{five}\:{other}\:{letters} \\ $$$${and}\:{then}\:{the}\:{three}\:{I}'{s}\:{in}\:\mathrm{6}\:{positions}: \\ $$$$\Box\mathrm{V}\Box\mathrm{S}\Box\mathrm{T}\Box\mathrm{N}\Box\mathrm{G}\Box \\ $$$$\mathrm{5}!×{C}_{\mathrm{3}} ^{\mathrm{6}} =\mathrm{2400} \\ $$$$ \\ $$$${Method}\:{II} \\ $$$${arrange}\:{at}\:{first}\:{the}\:{three}\:{I}'{s}\:{and} \\ $$$${then}\:{the}\:{five}\:{other}\:{letters}: \\ $$$$\Box\mathrm{I}\blacksquare\mathrm{I}\blacksquare\mathrm{I}\Box \\ $$$$\Box\:={zero}\:{or}\:{more}\:{other}\:{letters} \\ $$$$\blacksquare\:={one}\:{or}\:{more}\:{other}\:{letters} \\ $$$$\left(\underset{\Box} {\mathrm{1}+{x}+{x}^{\mathrm{2}} +…}\right)^{\mathrm{2}} \left(\underset{\blacksquare} {{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…}\right)^{\mathrm{2}} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$={x}^{\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{term}\:{x}^{\mathrm{5}} : \\ $$$${k}=\mathrm{3}\:\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{6}} \\ $$$$\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{5}!=\mathrm{2400} \\ $$
Commented by john_santu last updated on 06/Dec/20
$${great}\:{method}\:{II} \\ $$
Commented by malwan last updated on 06/Dec/20
$${I}\:{want}\:{to}\:{master}\:{method}\:{II} \\ $$$${like}\:{you}\:{mrW} \\ $$$${thank}\:{you} \\ $$
Commented by mr W last updated on 06/Dec/20
$${it}'{s}\:{a}\:{very}\:{powerful}\:{method},\:{try}\:{to} \\ $$$${apply}\:{it}! \\ $$
Answered by john_santu last updated on 06/Dec/20
Commented by bramlexs22 last updated on 06/Dec/20
$${great}! \\ $$