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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-




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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