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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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