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Question Number 125208 by liberty last updated on 09/Dec/20

There are 12 students in a party. Five of them are girls. In how many ways can these 12 students be arranged in a row if (i) there are no restrictions? (ii) the 5 girls must be together (forming a block)? (iii) no 2 girls are adjacent (iv) between two particular boys A and B , there no boys but exactly 3 girls?

$${There}\:{are}\:\mathrm{12}\:{students}\:{in}\:{a}\:{party}.\:{Five}\:{of} \\ $$$${them}\:{are}\:{girls}.\:{In}\:{how}\:{many}\:{ways}\:{can}\: \\ $$$${these}\:\mathrm{12}\:{students}\:{be}\:{arranged}\:{in}\:{a}\:{row}\:{if}\: \\ $$$$\left({i}\right)\:{there}\:{are}\:{no}\:{restrictions}? \\ $$$$\left({ii}\right)\:{the}\:\mathrm{5}\:{girls}\:{must}\:{be}\:{together}\:\left({forming}\:{a}\:{block}\right)? \\ $$$$\left({iii}\right)\:{no}\:\mathrm{2}\:{girls}\:{are}\:{adjacent}\: \\ $$$$\left({iv}\right)\:{between}\:{two}\:{particular}\:{boys}\:{A}\:{and}\:{B}\: \\ $$$$\:,\:{there}\:{no}\:{boys}\:{but}\:{exactly}\:\mathrm{3}\:{girls}? \\ $$

Answered by john_santu last updated on 09/Dec/20

(i) 12! (ii) 5! (7+1)! = 5! 8! (iii) −_1 G_1 −_2 ^x G_2 −_3 ^x G_3 −_4 ^x G_4 −_5 ^x G_5 −_6 = 5! 7! (((6+3−1)),(( 3)) ) = 5! 7! ((8),(3) ) (iv) AGGGB−−−−−−− −AGGGB−−−−−− −−AGGGB−−−−− −−−AGGGB−−−− −−−−AGGGB−−− −−−−−AGGGB−− −−−−−−AGGGB− −−−−−−−AGGGB the number of ways is = 2×8×C_3 ^( 5) ×3!×7! =12×8! ×10 = 120×8!

$$\left({i}\right)\:\mathrm{12}!\: \\ $$$$\left({ii}\right)\:\mathrm{5}!\:\left(\mathrm{7}+\mathrm{1}\right)!\:=\:\mathrm{5}!\:\mathrm{8}! \\ $$$$\left({iii}\right)\:\underset{\mathrm{1}} {−}\:{G}_{\mathrm{1}} \:\underset{\mathrm{2}} {\overset{{x}} {−}}\:{G}_{\mathrm{2}} \:\underset{\mathrm{3}} {\overset{{x}} {−}}{G}_{\mathrm{3}} \:\underset{\mathrm{4}} {\overset{{x}} {−}}\:{G}_{\mathrm{4}} \:\underset{\mathrm{5}} {\overset{{x}} {−}}\:{G}_{\mathrm{5}} \:\underset{\mathrm{6}} {−} \\ $$$$\:=\:\mathrm{5}!\:\mathrm{7}!\:\begin{pmatrix}{\mathrm{6}+\mathrm{3}−\mathrm{1}}\\{\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:=\:\mathrm{5}!\:\mathrm{7}!\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{3}}\end{pmatrix} \\ $$$$\left({iv}\right)\:{AGGGB}−−−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:−{AGGGB}−−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:−−{AGGGB}−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−{AGGGB}−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−−{AGGGB}−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−−−{AGGGB}−− \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−−−−{AGGGB}− \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−−−−−{AGGGB} \\ $$$$\:{the}\:{number}\:{of}\:{ways}\:{is}\: \\ $$$$\:=\:\mathrm{2}×\mathrm{8}×{C}_{\mathrm{3}} ^{\:\mathrm{5}} ×\mathrm{3}!×\mathrm{7}! \\ $$$$\:=\mathrm{12}×\mathrm{8}!\:×\mathrm{10}\:=\:\mathrm{120}×\mathrm{8}! \\ $$

Commented by liberty last updated on 09/Dec/20

yes..correct..thanks

$${yes}..{correct}..{thanks} \\ $$

Commented by malwan last updated on 16/Jan/21

please I can′t understand number(iii) (((6+3−1)),(3) ) what is it?6? 3? if girls =3 and boys=6 then what is the solution? and what if girls=n and boys=m ; m>n?

$${please} \\ $$$${I}\:{can}'{t}\:{understand}\:{number}\left({iii}\right) \\ $$$$\begin{pmatrix}{\mathrm{6}+\mathrm{3}−\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:{what}\:{is}\:{it}?\mathrm{6}?\:\mathrm{3}? \\ $$$${if}\:{girls}\:=\mathrm{3}\:{and}\:{boys}=\mathrm{6}\:{then} \\ $$$${what}\:{is}\:{the}\:{solution}? \\ $$$${and}\:{what}\:{if}\:{girls}={n}\:{and} \\ $$$${boys}={m}\:;\:{m}>{n}? \\ $$

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