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Let-n-be-the-number-of-ways-in-which-5-boys-and-5-girls-stand-in-a-queue-in-such-a-way-that-all-the-girls-stand-consecutively-in-the-queue-Let-m-be-the-number-of-ways-in-which-5-boys-and-5-girls-can-

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Question Number 21977 by Tinkutara last updated on 08/Oct/17
Let n be the number of ways in which 5 boys and 5 girls stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of (m/n) is
$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{in}\:\mathrm{which} \\ $$$$\mathrm{5}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{5}\:\mathrm{girls}\:\mathrm{stand}\:\mathrm{in}\:\mathrm{a}\:\mathrm{queue}\:\mathrm{in} \\ $$$$\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{all}\:\mathrm{the}\:\mathrm{girls}\:\mathrm{stand} \\ $$$$\mathrm{consecutively}\:\mathrm{in}\:\mathrm{the}\:\mathrm{queue}.\:\mathrm{Let}\:\mathrm{m}\:\mathrm{be} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{in}\:\mathrm{which}\:\mathrm{5}\:\mathrm{boys} \\ $$$$\mathrm{and}\:\mathrm{5}\:\mathrm{girls}\:\mathrm{can}\:\mathrm{stand}\:\mathrm{in}\:\mathrm{a}\:\mathrm{queue}\:\mathrm{in}\:\mathrm{such} \\ $$$$\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{exactly}\:\mathrm{four}\:\mathrm{girls}\:\mathrm{stand} \\ $$$$\mathrm{consecutively}\:\mathrm{in}\:\mathrm{the}\:\mathrm{queue}.\:\mathrm{Then}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\frac{{m}}{{n}}\:\mathrm{is} \\ $$
Commented by mrW1 last updated on 09/Oct/17
n=6!×5! (=86400) m=C_1 ^5 ×7!×4!−2×6!×5! (=432000) [or m=C_1 ^5 ×(7!−2!×6!)×4!=432000] ⇒(m/n)=((C_1 ^5 ×7!×4!)/(6!×5!))−2=7−2=5
$$\mathrm{n}=\mathrm{6}!×\mathrm{5}!\:\left(=\mathrm{86400}\right) \\ $$$$\mathrm{m}=\mathrm{C}_{\mathrm{1}} ^{\mathrm{5}} ×\mathrm{7}!×\mathrm{4}!−\mathrm{2}×\mathrm{6}!×\mathrm{5}!\:\left(=\mathrm{432000}\right) \\ $$$$\left[\mathrm{or}\:\mathrm{m}=\mathrm{C}_{\mathrm{1}} ^{\mathrm{5}} ×\left(\mathrm{7}!−\mathrm{2}!×\mathrm{6}!\right)×\mathrm{4}!=\mathrm{432000}\right] \\ $$$$\Rightarrow\frac{\mathrm{m}}{\mathrm{n}}=\frac{\mathrm{C}_{\mathrm{1}} ^{\mathrm{5}} ×\mathrm{7}!×\mathrm{4}!}{\mathrm{6}!×\mathrm{5}!}−\mathrm{2}=\mathrm{7}−\mathrm{2}=\mathrm{5} \\ $$
Commented by Tinkutara last updated on 09/Oct/17
6 is wrong.
$$\mathrm{6}\:\mathrm{is}\:\mathrm{wrong}. \\ $$
Commented by mrW1 last updated on 09/Oct/17
I have found the mistake. see correction.
$$\mathrm{I}\:\mathrm{have}\:\mathrm{found}\:\mathrm{the}\:\mathrm{mistake}.\:\mathrm{see}\:\mathrm{correction}. \\ $$
Commented by Tinkutara last updated on 09/Oct/17
Thank you very much Sir! It includes permutations also.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}!\:\mathrm{It}\:\mathrm{includes} \\ $$$$\mathrm{permutations}\:\mathrm{also}. \\ $$
Commented by Rasheed.Sindhi last updated on 09/Oct/17
Sir mrW1 Please write your answer in some detail in order to make it easy to understand.Thanks in advance.
$$\mathrm{Sir}\:\mathrm{mrW1} \\ $$$$\mathrm{Please}\:\mathrm{write}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{in} \\ $$$$\mathrm{some}\:\mathrm{detail}\:\mathrm{in}\:\mathrm{order}\:\mathrm{to}\:\mathrm{make} \\ $$$$\mathrm{it}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{understand}.\mathrm{Thanks} \\ $$$$\mathrm{in}\:\mathrm{advance}. \\ $$
Commented by mrW1 last updated on 09/Oct/17
case 1: BBB[GGGGG]BB since the five girls are together, we treat them as a single object. to arrange this object and the five boys there are 6! ways. to arrange the five girls in the object there are 5! ways. ⇒n=6!5! case 2: BGBB[GGGG]BB we divide the five girls into a single person G and a groupe with 4 persons [GGGG]. there are C_1 ^5 ways to do this. to arrange the four girls in the groupe there are 4! ways. to arrange the five boys and the single girl and the groupe there are 7! ways. ⇒C_1 ^5 ×4!×7! but in these arrangements we have also G[GGGG] and [GGGG]G, both mean that five girls are together. these arrangements must be excluded. ⇒m=C_1 ^5 ×4!×7!−2n ⇒m=C_1 ^5 ×4!×7!−2×6!×5!
$$\mathrm{case}\:\mathrm{1}: \\ $$$$\mathrm{BBB}\left[\mathrm{GGGGG}\right]\mathrm{BB} \\ $$$$\mathrm{since}\:\mathrm{the}\:\mathrm{five}\:\mathrm{girls}\:\mathrm{are}\:\mathrm{together},\:\mathrm{we} \\ $$$$\mathrm{treat}\:\mathrm{them}\:\mathrm{as}\:\mathrm{a}\:\mathrm{single}\:\mathrm{object}.\:\mathrm{to}\:\mathrm{arrange} \\ $$$$\mathrm{this}\:\mathrm{object}\:\mathrm{and}\:\mathrm{the}\:\mathrm{five}\:\mathrm{boys}\:\mathrm{there}\:\mathrm{are} \\ $$$$\mathrm{6}!\:\mathrm{ways}.\:\mathrm{to}\:\mathrm{arrange}\:\mathrm{the}\:\mathrm{five}\:\mathrm{girls}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{object}\:\mathrm{there}\:\mathrm{are}\:\mathrm{5}!\:\mathrm{ways}. \\ $$$$\Rightarrow\mathrm{n}=\mathrm{6}!\mathrm{5}! \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2}: \\ $$$$\mathrm{BGBB}\left[\mathrm{GGGG}\right]\mathrm{BB} \\ $$$$\mathrm{we}\:\mathrm{divide}\:\mathrm{the}\:\mathrm{five}\:\mathrm{girls}\:\mathrm{into}\:\mathrm{a}\:\mathrm{single} \\ $$$$\mathrm{person}\:\mathrm{G}\:\mathrm{and}\:\mathrm{a}\:\mathrm{groupe}\:\mathrm{with}\:\mathrm{4}\:\mathrm{persons} \\ $$$$\left[\mathrm{GGGG}\right].\:\mathrm{there}\:\mathrm{are}\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{5}} \:\mathrm{ways}\:\mathrm{to}\:\mathrm{do}\:\mathrm{this}. \\ $$$$\mathrm{to}\:\mathrm{arrange}\:\mathrm{the}\:\mathrm{four}\:\mathrm{girls}\:\mathrm{in}\:\mathrm{the}\:\mathrm{groupe} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{4}!\:\mathrm{ways}. \\ $$$$\mathrm{to}\:\mathrm{arrange}\:\mathrm{the}\:\mathrm{five}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{the}\:\mathrm{single} \\ $$$$\mathrm{girl}\:\mathrm{and}\:\mathrm{the}\:\mathrm{groupe}\:\mathrm{there}\:\mathrm{are}\:\mathrm{7}!\:\mathrm{ways}. \\ $$$$\Rightarrow\mathrm{C}_{\mathrm{1}} ^{\mathrm{5}} ×\mathrm{4}!×\mathrm{7}! \\ $$$$\mathrm{but}\:\mathrm{in}\:\mathrm{these}\:\mathrm{arrangements}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{also}\:\mathrm{G}\left[\mathrm{GGGG}\right]\:\mathrm{and}\:\left[\mathrm{GGGG}\right]\mathrm{G},\:\mathrm{both} \\ $$$$\mathrm{mean}\:\mathrm{that}\:\mathrm{five}\:\mathrm{girls}\:\mathrm{are}\:\mathrm{together}.\:\mathrm{these} \\ $$$$\mathrm{arrangements}\:\mathrm{must}\:\mathrm{be}\:\mathrm{excluded}. \\ $$$$\Rightarrow\mathrm{m}=\mathrm{C}_{\mathrm{1}} ^{\mathrm{5}} ×\mathrm{4}!×\mathrm{7}!−\mathrm{2n} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{C}_{\mathrm{1}} ^{\mathrm{5}} ×\mathrm{4}!×\mathrm{7}!−\mathrm{2}×\mathrm{6}!×\mathrm{5}! \\ $$
Commented by Rasheed.Sindhi last updated on 09/Oct/17
Thanks again!
$$\mathrm{Thanks}\:\mathrm{again}! \\ $$
Commented by mrW1 last updated on 09/Oct/17
you are welcome sir!
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir}! \\ $$
Answered by Rasheed.Sindhi last updated on 08/Oct/17
Number of places for 5 girls among boys represented by(−): − b_1 − b_2 − b_3 − b_4 − b_5 − n=6 Number of places for 4 girls=6 Number of places for 5th girl for each of 6 cases of 4 girls=5 ∴ m=6×5=30 (m/n)=((30)/6)=5
$$\mathrm{Number}\:\mathrm{of}\:\mathrm{places}\:\mathrm{for}\:\mathrm{5}\:\mathrm{girls} \\ $$$$\mathrm{among}\:\mathrm{boys}\:\mathrm{represented}\:\mathrm{by}\left(−\right): \\ $$$$\:\:\:−\:\mathrm{b}_{\mathrm{1}} −\:\mathrm{b}_{\mathrm{2}} −\:\mathrm{b}_{\mathrm{3}} −\:\mathrm{b}_{\mathrm{4}} −\:\mathrm{b}_{\mathrm{5}} − \\ $$$$\mathrm{n}=\mathrm{6} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{places}\:\mathrm{for}\:\mathrm{4}\:\mathrm{girls}=\mathrm{6} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{places}\:\mathrm{for}\:\mathrm{5th}\:\mathrm{girl} \\ $$$$\mathrm{for}\:\mathrm{each}\:\mathrm{of}\:\mathrm{6}\:\mathrm{cases}\:\mathrm{of}\:\mathrm{4}\:\mathrm{girls}=\mathrm{5} \\ $$$$\therefore\:\:\:\:\:\mathrm{m}=\mathrm{6}×\mathrm{5}=\mathrm{30} \\ $$$$\:\:\:\:\:\frac{\mathrm{m}}{\mathrm{n}}=\frac{\mathrm{30}}{\mathrm{6}}=\mathrm{5} \\ $$
Commented by Rasheed.Sindhi last updated on 08/Oct/17
I didn′t consider order between girls themselves or boys themselves. I considerd order only between boys and girls. Perhaps I′m wrong.
$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{consider}\:\mathrm{order}\:\mathrm{between} \\ $$$$\mathrm{girls}\:\mathrm{themselves}\:\:\mathrm{or}\:\mathrm{boys}\:\mathrm{themselves}. \\ $$$$\mathrm{I}\:\mathrm{considerd}\:\mathrm{order}\:\mathrm{only}\:\mathrm{between}\:\mathrm{boys} \\ $$$$\mathrm{and}\:\mathrm{girls}.\:\mathrm{Perhaps}\:\mathrm{I}'\mathrm{m}\:\mathrm{wrong}. \\ $$
Commented by Rasheed.Sindhi last updated on 08/Oct/17
I think that the wording of the question doesn′t demand order between girls when it says “... in such a way that all the girls stand consecutively in a queue.” This means order is not matter. Only all the girls shoud stand without any boy/s between them.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{the}\:\mathrm{wording}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{question}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{demand}\:\mathrm{order} \\ $$$$\mathrm{between}\:\mathrm{girls}\:\mathrm{when}\:\mathrm{it}\:\mathrm{says}\:“… \\ $$$$\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{all}\:\mathrm{the}\:\mathrm{girls} \\ $$$$\mathrm{stand}\:\mathrm{consecutively}\:\mathrm{in}\:\mathrm{a}\:\mathrm{queue}.'' \\ $$$$\mathrm{This}\:\mathrm{means}\:\mathrm{order}\:\mathrm{is}\:\mathrm{not}\:\mathrm{matter}. \\ $$$$\mathrm{Only}\:\mathrm{all}\:\mathrm{the}\:\mathrm{girls}\:\mathrm{shoud}\:\mathrm{stand} \\ $$$$\mathrm{without}\:\mathrm{any}\:\mathrm{boy}/\mathrm{s}\:\mathrm{between}\:\mathrm{them}. \\ $$
Commented by Tinkutara last updated on 09/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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