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Question Number 121851 by bemath last updated on 12/Nov/20

$$Howmanywaystoarrangeletters$$$$usinglettersinthewordOKINKO$$$$\left(a\right)1260\left(b\right)1160\left(c\right)980$$$$\left(d\right)880\left(e\right)720$$

Commented by liberty last updated on 12/Nov/20

$$\mathrm{case}\left(1\right):1-\mathrm{alphabet}\Rightarrow {\mathrm{C}}_1^4=4$$$$\mathrm{case}\left(2\right):2-\mathrm{alphabet}\to \{\begin{array}{l}\mathrm{OO},\mathrm{KK}=2\\ \mathrm{OK},\mathrm{KO}=2\\ \mathrm{OI},\mathrm{ON},\mathrm{KI},\mathrm{KN},\mathrm{IN}=2!\times 5=10\end{array}$$$$\mathrm{case}\left(3\right):3-\mathrm{aphabet}$$$$\to \{\begin{array}{l}\mathrm{OOK}=\frac{3!}{2!}=3,\mathrm{OKK}=\frac{3!}{2}=3\\ \mathrm{OO}\to \{\begin{array}{l}\mathrm{I}\\ \mathrm{N}\end{array}=2\times \frac{3!}{2!}=6;\mathrm{KK}\to \{\begin{array}{l}\mathrm{I}\\ \mathrm{N}\end{array}=2\times \frac{3!}{2!}=6\\ \mathrm{OK}\to \{\begin{array}{l}\mathrm{I}\\ \mathrm{N}\end{array}=2\times 3!=12\end{array}$$$$\mathrm{case}\left(4\right):4-\mathrm{alphabet}$$$$\to \{\begin{array}{l}\mathrm{OOKK}=\frac{4!}{2!.2!}=6\\ \mathrm{OOK}\{\begin{array}{l}\mathrm{I}\\ \mathrm{N}\end{array}=2\times \frac{4!}{2!}=24\\ \mathrm{OKK}\{\begin{array}{l}\mathrm{I}\\ \mathrm{N}\end{array}=2\times \frac{4!}{2!}=24\\ \mathrm{OKIN}=4!=24\end{array}$$$$\mathrm{case}\left(5\right):5-\mathrm{alphabet}$$$$\to \{\begin{array}{l}\mathrm{OOKK}\{\begin{array}{l}\mathrm{I}\\ \mathrm{N}\end{array}=2\times \frac{5!}{2!.2!}=60\\ \mathrm{OKKNI}=\frac{5!}{2!}=60\\ \mathrm{OOKNI}=\frac{5!}{2!}=60\end{array}$$$$\mathrm{case}\left(6\right):6-\mathrm{alphabet}=\frac{6!}{2!.2!}=180$$$$\mathrm{Totally}=18+30+78+180+180$$$$360+108+18$$$$486.0$$

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