How-many-4-letters-word-can-be-formed-from-COMMUNICATION-

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

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

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

2×C 2×O 2×M 2×N 2×I 1×U 1×A 1×T case 1: WXYZ 4!(1+x)^5 (1+x)^3 =...1680x^4 ... or C_4 ^8 ×4!=1680 case 2: XYYZ ((4!)/(2!))×C_1 ^5 ×x^2 (1+x)^4 (1+x)^3 =...1260x^4 ... or C_1 ^5 ×C_2 ^7 ×((4!)/(2!))=1260 case 3: XXYY ((4!)/(2!2!))×C_2 ^5 ×x^2 ×x^2 (1+x)^3 (1+x)^3 =...60x^4 ... or C_2 ^5 ×((4!)/(2!2!))=60 total 1680+1260+60=3000 words

$$\mathrm{2}×{C} \\ $$$$\mathrm{2}×{O} \\ $$$$\mathrm{2}×{M} \\ $$$$\mathrm{2}×{N} \\ $$$$\mathrm{2}×{I} \\ $$$$\mathrm{1}×{U} \\ $$$$\mathrm{1}×{A} \\ $$$$\mathrm{1}×{T} \\ $$$$ \\ $$$${case}\:\mathrm{1}:\:{WXYZ} \\ $$$$\mathrm{4}!\left(\mathrm{1}+{x}\right)^{\mathrm{5}} \left(\mathrm{1}+{x}\right)^{\mathrm{3}} =…\mathrm{1680}{x}^{\mathrm{4}} … \\ $$$${or}\:{C}_{\mathrm{4}} ^{\mathrm{8}} ×\mathrm{4}!=\mathrm{1680} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{XYYZ} \\ $$$$\frac{\mathrm{4}!}{\mathrm{2}!}×{C}_{\mathrm{1}} ^{\mathrm{5}} ×{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{4}} \left(\mathrm{1}+{x}\right)^{\mathrm{3}} =…\mathrm{1260}{x}^{\mathrm{4}} … \\ $$$${or}\:{C}_{\mathrm{1}} ^{\mathrm{5}} ×{C}_{\mathrm{2}} ^{\mathrm{7}} ×\frac{\mathrm{4}!}{\mathrm{2}!}=\mathrm{1260} \\ $$$$ \\ $$$${case}\:\mathrm{3}:\:{XXYY} \\ $$$$\frac{\mathrm{4}!}{\mathrm{2}!\mathrm{2}!}×{C}_{\mathrm{2}} ^{\mathrm{5}} ×{x}^{\mathrm{2}} ×{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{3}} \left(\mathrm{1}+{x}\right)^{\mathrm{3}} =…\mathrm{60}{x}^{\mathrm{4}} … \\ $$$${or}\:{C}_{\mathrm{2}} ^{\mathrm{5}} ×\frac{\mathrm{4}!}{\mathrm{2}!\mathrm{2}!}=\mathrm{60} \\ $$$$ \\ $$$${total}\:\mathrm{1680}+\mathrm{1260}+\mathrm{60}=\mathrm{3000}\:{words} \\ $$

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

Sir that is coefficient of x^4 in 4!(1 + (x/(1!)) + (x^2 /(2!)))^5 (1 + x)^3 Thanks sir

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

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