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Question-31591




Question Number 31591 by Tinkutara last updated on 10/Mar/18
Commented by mrW2 last updated on 11/Mar/18
no answer seems to be correct
$${no}\:{answer}\:{seems}\:{to}\:{be}\:{correct} \\ $$
Commented by Tinkutara last updated on 11/Mar/18
Yes I also think. Does it should be  ^7 P_4 ×6! ?
$${Yes}\:{I}\:{also}\:{think}.\:{Does}\:{it}\:{should}\:{be} \\ $$$$\:^{\mathrm{7}} {P}_{\mathrm{4}} ×\mathrm{6}!\:? \\ $$
Commented by mrW2 last updated on 11/Mar/18
to arrange the 6 boys:  6! ways  to arrange 4 girls into 7 positions:  P_4 ^( 7)  ways    ⇒6!×P_4 ^( 7) =840×6!=120×7!
$${to}\:{arrange}\:{the}\:\mathrm{6}\:{boys}: \\ $$$$\mathrm{6}!\:{ways} \\ $$$${to}\:{arrange}\:\mathrm{4}\:{girls}\:{into}\:\mathrm{7}\:{positions}: \\ $$$${P}_{\mathrm{4}} ^{\:\mathrm{7}} \:{ways} \\ $$$$ \\ $$$$\Rightarrow\mathrm{6}!×{P}_{\mathrm{4}} ^{\:\mathrm{7}} =\mathrm{840}×\mathrm{6}!=\mathrm{120}×\mathrm{7}! \\ $$
Commented by Tinkutara last updated on 11/Mar/18
Thank you very much Sir! I got the answer. ������☺
Answered by MJS last updated on 11/Mar/18
the girls can be on these positions  (0 stands for 10):  1357 1358 1359 1350  1368 1369 1360  1379 1370  1380  1468 1469 1460  1479 1470  1480  1579 1570  1580  1680  2468 2469 2460  2479 2470  2480  2579 2570  2580  2680  3579 3570  3580  3680  4680  so it′s 35×4!×6!=5×7×4!×6!=  =5!×7!
$$\mathrm{the}\:\mathrm{girls}\:\mathrm{can}\:\mathrm{be}\:\mathrm{on}\:\mathrm{these}\:\mathrm{positions} \\ $$$$\left(\mathrm{0}\:\mathrm{stands}\:\mathrm{for}\:\mathrm{10}\right): \\ $$$$\mathrm{1357}\:\mathrm{1358}\:\mathrm{1359}\:\mathrm{1350} \\ $$$$\mathrm{1368}\:\mathrm{1369}\:\mathrm{1360} \\ $$$$\mathrm{1379}\:\mathrm{1370} \\ $$$$\mathrm{1380} \\ $$$$\mathrm{1468}\:\mathrm{1469}\:\mathrm{1460} \\ $$$$\mathrm{1479}\:\mathrm{1470} \\ $$$$\mathrm{1480} \\ $$$$\mathrm{1579}\:\mathrm{1570} \\ $$$$\mathrm{1580} \\ $$$$\mathrm{1680} \\ $$$$\mathrm{2468}\:\mathrm{2469}\:\mathrm{2460} \\ $$$$\mathrm{2479}\:\mathrm{2470} \\ $$$$\mathrm{2480} \\ $$$$\mathrm{2579}\:\mathrm{2570} \\ $$$$\mathrm{2580} \\ $$$$\mathrm{2680} \\ $$$$\mathrm{3579}\:\mathrm{3570} \\ $$$$\mathrm{3580} \\ $$$$\mathrm{3680} \\ $$$$\mathrm{4680} \\ $$$$\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\mathrm{35}×\mathrm{4}!×\mathrm{6}!=\mathrm{5}×\mathrm{7}×\mathrm{4}!×\mathrm{6}!= \\ $$$$=\mathrm{5}!×\mathrm{7}! \\ $$
Commented by mrW2 last updated on 11/Mar/18
this is my solution:  between two girls there must be at  least one boy, that′s to say the four  girls can only be placed in the positions  marked as ⊝:  ⊝⊛⊝⊛⊝⊛⊝⊛⊝⊛⊝⊛⊝  where ⊛ means a boy.    following is e.g. a valid arrangement:  ⊝⊛□⊛⊝⊛□⊛□⊛⊝⊛□  where □ means a girl, ⊝ means empty.    to arrange the 6 boys ⊛: 6! ways  to arrange the 4 girls □: P_4 ^( 7)  ways  ⇒result=P_4 ^( 7) ×6! ways.    generally: n students, m girls  ⇒P_m ^( n−m+1) ×(n−m)!
$${this}\:{is}\:{my}\:{solution}: \\ $$$${between}\:{two}\:{girls}\:{there}\:{must}\:{be}\:{at} \\ $$$${least}\:{one}\:{boy},\:{that}'{s}\:{to}\:{say}\:{the}\:{four} \\ $$$${girls}\:{can}\:{only}\:{be}\:{placed}\:{in}\:{the}\:{positions} \\ $$$${marked}\:{as}\:\circleddash: \\ $$$$\circleddash\circledast\circleddash\circledast\circleddash\circledast\circleddash\circledast\circleddash\circledast\circleddash\circledast\circleddash \\ $$$${where}\:\circledast\:{means}\:{a}\:{boy}. \\ $$$$ \\ $$$${following}\:{is}\:{e}.{g}.\:{a}\:{valid}\:{arrangement}: \\ $$$$\circleddash\circledast\square\circledast\circleddash\circledast\square\circledast\square\circledast\circleddash\circledast\square \\ $$$${where}\:\square\:{means}\:{a}\:{girl},\:\circleddash\:{means}\:{empty}. \\ $$$$ \\ $$$${to}\:{arrange}\:{the}\:\mathrm{6}\:{boys}\:\circledast:\:\mathrm{6}!\:{ways} \\ $$$${to}\:{arrange}\:{the}\:\mathrm{4}\:{girls}\:\square:\:{P}_{\mathrm{4}} ^{\:\mathrm{7}} \:{ways} \\ $$$$\Rightarrow{result}={P}_{\mathrm{4}} ^{\:\mathrm{7}} ×\mathrm{6}!\:{ways}. \\ $$$$ \\ $$$${generally}:\:{n}\:{students},\:{m}\:{girls} \\ $$$$\Rightarrow{P}_{{m}} ^{\:{n}−{m}+\mathrm{1}} ×\left({n}−{m}\right)! \\ $$

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