along with group :( - permutation / combination prob

lets count each couple as 1 so there are total of 6 people (each couple counted as 1 each)
6 people can sit in 6! ways
now each couple can sit in 2 ways * 2 ways
total ways = 6!*2*2 = 2880

AIM GMAT wrote:A group of 8 students take up a full row at a movie theater. There are 2 couples and the
2 people within each couple must sit together. In how many di& erent arrangements can
the students sit?

let c1 represents couple 1, and c2 represents couple 2, and the remaining four students be represented as s1,s2,s3,s4. Also let people within c1 and c2 be p1,p2,p3,p4 respectively.

as two persons within each couple must sit together therefore people within c1 and c2 will be consider as 1, so therefore we have total of 6 objects. these six objects can arrange themselves in 6! ways.

c1[(p1,p2)],c2[(p3,p4)],s1,s2,s3,s4 {it is one of the way of arrangement.}

now p1,p2 can interchange their position within a group in 2! ways similarly for p3 and p4 we have 2! ways.

so total no. of ways of arrangements would be 6!*2!*2!=2880

Total ways = 6!*2!*2! = 2880

Since the 2 couples must sit together, consider them as 1 person to start with, so there are 6 students, which can sit in
6! ways,

Now there were 2 couples, which we substituted as 1 student each to find the unique ways, get them back
6! x 2! x 2!

Is this the correct answer ?

Thank You,

Yep guys you all are correct , the answer is 2880.