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Question Number 130019 by I want to learn more last updated on 21/Jan/21

$$\mathrm{How}\mathrm{many}4\mathrm{letters}\mathrm{word}\mathrm{can}\mathrm{be}\mathrm{formed}\mathrm{from}\mathrm{COMMUNICATION}$$

Answered by mr W last updated on 22/Jan/21

$$2\times C$$$$2\times O$$$$2\times M$$$$2\times N$$$$2\times I$$$$1\times U$$$$1\times A$$$$1\times T$$$$$$$$case1:WXYZ$$$$4!{(1+x)}^5{(1+x)}^3=...1680x^4...$$$$or{C}_4^8\times 4!=1680$$$$$$$$case2:XYYZ$$$$\frac{4!}{2!}\times C_1^5\times x^2{(1+x)}^4{(1+x)}^3=...1260x^4...$$$$or{C}_1^5\times C_2^7\times \frac{4!}{2!}=1260$$$$$$$$case3:XXYY$$$$\frac{4!}{2!2!}\times C_2^5\times x^2\times x^2{(1+x)}^3{(1+x)}^3=...60x^4...$$$$or{C}_2^5\times \frac{4!}{2!2!}=60$$$$$$$$total1680+1260+60=3000words$$

Commented by I want to learn more last updated on 22/Jan/21

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

Commented by I want to learn more last updated on 23/Jan/21

$$\mathrm{Sir}\mathrm{that}\mathrm{is}\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^4\mathrm{in}4!{(1+\frac{\mathrm{x}}{1!}+\frac{{\mathrm{x}}^2}{2!})}^5{(1+\mathrm{x})}^3$$$$\mathrm{Thanks}\mathrm{sir}$$

Commented by mr W last updated on 23/Jan/21

$$yes.$$

Commented by I want to learn more last updated on 23/Jan/21

$$\mathrm{Thanks}\mathrm{sir}.$$

Commented by mr W last updated on 24/Jan/21

$$seealsoQ130193$$

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