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Question Number 10387 by FilupSmith last updated on 06/Feb/17

6 people a, b, c, d, e, and f stand in a line.     The number of ways they can stand arranged  is equal to 6!     If two people have to stand next to each other,  but everyone else do not matter, how many combinations  combinations are there?     e.g.   (ab)cdef    or    cd(ba)ef

$$\mathrm{6}\:\mathrm{people}\:{a},\:{b},\:{c},\:{d},\:{e},\:\mathrm{and}\:{f}\:\mathrm{stand}\:\mathrm{in}\:\mathrm{a}\:\mathrm{line}. \\ $$$$\: \\ $$$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{they}\:\mathrm{can}\:\mathrm{stand}\:\mathrm{arranged} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{6}! \\ $$$$\: \\ $$$$\mathrm{If}\:\mathrm{two}\:\mathrm{people}\:\mathrm{have}\:\mathrm{to}\:\mathrm{stand}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}, \\ $$$$\mathrm{but}\:\mathrm{everyone}\:\mathrm{else}\:\mathrm{do}\:\mathrm{not}\:\mathrm{matter},\:\mathrm{how}\:\mathrm{many}\:\mathrm{combinations} \\ $$$$\mathrm{combinations}\:\mathrm{are}\:\mathrm{there}? \\ $$$$\: \\ $$$$\mathrm{e}.\mathrm{g}.\:\:\:\left({ab}\right){cdef}\:\:\:\:\mathrm{or}\:\:\:\:{cd}\left({ba}\right){ef} \\ $$

Answered by sandy_suhendra last updated on 06/Feb/17

the two people we assume as 1 people  so there are 5 people ⇒ the combination=5!  but the two people have combination =2!  the total combination=2!×5!=240

$$\mathrm{the}\:\mathrm{two}\:\mathrm{people}\:\mathrm{we}\:\mathrm{assume}\:\mathrm{as}\:\mathrm{1}\:\mathrm{people} \\ $$$$\mathrm{so}\:\mathrm{there}\:\mathrm{are}\:\mathrm{5}\:\mathrm{people}\:\Rightarrow\:\mathrm{the}\:\mathrm{combination}=\mathrm{5}! \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{two}\:\mathrm{people}\:\mathrm{have}\:\mathrm{combination}\:=\mathrm{2}! \\ $$$$\mathrm{the}\:\mathrm{total}\:\mathrm{combination}=\mathrm{2}!×\mathrm{5}!=\mathrm{240} \\ $$

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