Question

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

Answer 1

I am inclined to think:
\[2!^3 * 3! = 48 \]

Answer 2

you have BBGGAA

Answer 3

can we think in linear way??

Answer 4

what do you mean?

Answer 5

how did you get that answer

Answer 6

or kindely explain how u did?

Answer 7

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 = \[2! * 2! * 2!\]

Answer 8

but! it can be:
couple1 couple2 couple3
couple3 couple2 couple1

etc...

so we can arrange the 3 couples in 3! ways.

Answer 9

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

Answer 10

you're treating the two boys as a couple?

Answer 11

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).

Answer 12

but you can rearrange the couples

Answer 13

yes. you can arrange the couples between the three rows. thus the 3!

Answer 14

we have two rows
_ _ _
_ _ _

Answer 15

ah i am so wrong. you are right.

Answer 16

yes thats what m thinking .. how to solve its definitely not circular and I doubt on Linear permutation

Answer 17

thats why i got confused about your answer

Answer 18

heh sorry. i misread the question :(. that's what i get for skimming.

Answer 19

i will forgive you if you help me with my question, brb

Answer 20

but go ahead and try to answer

Answer 21

so you solved the question, 3 rows, boy cant sit next to girl?

Answer 22

so i guess is it correct??
B1 B2 A1 ----6 WAYS
G1 G2 A2 ----6 WAYS ie, TOTAL =36 ways

Answer 23

there is another set of cases:

B1 A1 G1
B2 A2 G2

Answer 24

B1 A1 G1 = 2 ways to arrange
B2 A2 G2 = 2 ways to arrange

Answer 25

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

Answer 26

12 * 2 * 2 (another 2 to choose between the A's) = 48 for that case.

Answer 27

times another 2 to account for being able to switch the rows: r1, r2 or r2, r1

Answer 28

= 96 for that case.

you'll have to + this to the possibilities for the other case

Answer 29

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?

Answer 30

I am also confused dear :)

Answer 31

128 maybe!

Answer 32

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

Answer 33

hi cantorset .. u ther??

Answer 34

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.

Answer 35

need more explanation?

Answer 36

yes i do, i got the same answer but i didnt get your 2x3x3x2! , etc

Answer 37

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.