anonymous
  • anonymous
There are two rows of seats with three side-by-side seats in each row. Two little boys, two little girls, and two adults sit in the six seats so that neither little boy sits to the side of either little girl. In how many different ways can these six people be seated? OPTIONS 1) 48 2) 128 3) 192 4) 176
Mathematics
chestercat
  • chestercat
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mattfeury
  • mattfeury
I am inclined to think: \[2!^3 * 3! = 48 \]
anonymous
  • anonymous
you have BBGGAA
anonymous
  • anonymous
can we think in linear way??

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mattfeury
  • mattfeury
what do you mean?
anonymous
  • anonymous
how did you get that answer
anonymous
  • anonymous
or kindely explain how u did?
mattfeury
  • mattfeury
if you a have of couple (2 people), you can arrange them in 2! ways. 2! = 2 * 1 = 2 if you have three couples in 3 rows, then they can be sat = \[2! * 2! * 2!\]
mattfeury
  • mattfeury
but! it can be: couple1 couple2 couple3 couple3 couple2 couple1 etc... so we can arrange the 3 couples in 3! ways.
mattfeury
  • mattfeury
so basically we have 6 seats. but first we simply that to three rows. we have three rows. a couple per row. there are 3! ways to arrange this. in each row, each couple can be arranged 2! ways. Since this is the case for each of the three rows, it is 2! * 2! * 2! multiplying all these possibilities together: 2! * 2! * 2! * 3! = 48
anonymous
  • anonymous
you're treating the two boys as a couple?
mattfeury
  • mattfeury
ah yeah sorry. i was thinking of a similar problem with 3 couples. it is the same principle though. 3 sets of 2 individuals (in this case: boys, girls, adults).
anonymous
  • anonymous
but you can rearrange the couples
mattfeury
  • mattfeury
yes. you can arrange the couples between the three rows. thus the 3!
anonymous
  • anonymous
we have two rows _ _ _ _ _ _
mattfeury
  • mattfeury
ah i am so wrong. you are right.
anonymous
  • anonymous
yes thats what m thinking .. how to solve its definitely not circular and I doubt on Linear permutation
anonymous
  • anonymous
thats why i got confused about your answer
mattfeury
  • mattfeury
heh sorry. i misread the question :(. that's what i get for skimming.
anonymous
  • anonymous
i will forgive you if you help me with my question, brb
anonymous
  • anonymous
but go ahead and try to answer
anonymous
  • anonymous
so you solved the question, 3 rows, boy cant sit next to girl?
anonymous
  • anonymous
so i guess is it correct?? B1 B2 A1 ----6 WAYS G1 G2 A2 ----6 WAYS ie, TOTAL =36 ways
mattfeury
  • mattfeury
there is another set of cases: B1 A1 G1 B2 A2 G2
mattfeury
  • mattfeury
B1 A1 G1 = 2 ways to arrange B2 A2 G2 = 2 ways to arrange
mattfeury
  • mattfeury
better: B1 A1 G1 = 4 * 3 to arrange (you have a set of {b1, b2, g1, g2} and you have 4 and 3 options for the end) B2 A2 G2 = 2 * 1 ways to arrange
mattfeury
  • mattfeury
12 * 2 * 2 (another 2 to choose between the A's) = 48 for that case.
mattfeury
  • mattfeury
times another 2 to account for being able to switch the rows: r1, r2 or r2, r1
mattfeury
  • mattfeury
= 96 for that case. you'll have to + this to the possibilities for the other case
mattfeury
  • mattfeury
i may be wrong though. i can't seem to figure out the other case to where it adds to an answer :( 96 + (2 * 2 * 2 * 2) = 112?
anonymous
  • anonymous
I am also confused dear :)
mattfeury
  • mattfeury
128 maybe!
anonymous
  • anonymous
can it be like? B A G B A G G A B G A B B B A G G A G G A B B A A B B A G G A G G A B B Each case can be arranged in 8 ways , i.e. Boys - 2! , Gals - 2! , adults - 2! Hence 6 * (2*2*2) = 48 Option 1
anonymous
  • anonymous
hi cantorset .. u ther??
nowhereman
  • nowhereman
The only possible cases are, if the girls sit on different rows, that the parents sit in the middle seats and the boys on the other sides. If the girls sit on the same row the third seat must be occupied by a parent and there is no further restriction on the other row. So you get for the number of constellations: In the first case 4*2!*2!*2! = 32 and in the second case 2*3*3*2!*2!*2! = 144, that means there are 176 possibilties.
nowhereman
  • nowhereman
need more explanation?
anonymous
  • anonymous
yes i do, i got the same answer but i didnt get your 2x3x3x2! , etc
nowhereman
  • nowhereman
2 because the girls can either sit in the first or in the second row 3 because you can choose any seat in that row for the parent 3 because you must choose one seat in the other row where the second parent sits 2!*2!*2! always because it doesn't matter if you interchange the 2 girls / boys / parents with each other.

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