In a row of 30 seats, six team of 4 persons each will be seated. The

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In a row of 30 seats, six team of 4 persons each will be seated. The members of any one team always sit together in four consecutive seats in the same order. Different teams may be adjacent or separated by empty seats. In how many ways can the six teams be seated in the 30 seats?

(A) (6!)(6!)

(B) \(\frac{12!}{6!}\)

(C) 12C6

(D) 30C12

(E) 30C6*6

chetan2u · Answer

Hi,
There are total 30 seats, but 4 members of 6 different groups always sit together and in same sequence...
So the group can be taken as one seat, making the total as 12 seats..

12 seats include 6 empty chairs as total 6*4 chairs are filled in 30 seats and these 6 seats will be identical while the other 6 would be different as each consists of different group..
Ans 12!/6!
B

OreoShake · Answer

Hi Chetan,

Im unable to undertand your explaination. Because the teams will have the same arrangement, we will have essentially 24/4 i.e 6 places+the remaining empty seats i.e 6; so the total becomes 12!. 12! is the answer according to me, please explain what is wrong.

thanks,
Jon

chetan2u · Answer

Hi Jon,

You are correct till 12! ...
But the 6 empty seats are same and this 6 seats can be arranged in 6! Ways ...
That is why we divide 12! By 6!

aa008 · Answer

say,team 1 =T1 and team2 =T2
like this we have T3,T4,T5,T6
EACH TEAM CONSISTS OF 4 MEMBERS, SO 6 SEATS ARE REMAINING EMPTY.
and each empty seat = E

ultimately question becomes in how many different way you can arrange T1,T2,T3,T4,T5,T6,E,E,E,E,E,E

=  12!/6!

note-  you do not need to multiple by 4! to arrange members of each team as question has clearly stated.

ScottTargetTestPrep · Answer

We can let the 6 teams be A, B, C, D, E and F and let S be an empty seat. Since there must be 6 empty seats, therefore, one seating arrangement is ABCDEFSSSSSS. Now we need to determine the number of ways we can arrange these 12 “letters.” By the formula for the permutation of indistinguishable objects, we see that these 12 “letters” can be arranged in 12! / 6! ways.

Answer: B

Kinshook · Answer

In a row of 30 seats, six team of 4 persons each will be seated. The members of any one team always sit together in four consecutive seats in the same order. Different teams may be adjacent or separated by empty seats.

In how many ways can the six teams be seated in the 30 seats?

There are 6 teams and 4 members in each team to be seated in the same order. There are 6! ways to arrange the teams. Therefore, there are 6! ways to seat them on 24 seats. But there are 6 vacant seats. Let us consider teams as a unit and a unit occupies 4 consecutive seats.

We have set of 6 4 consecutive seat pairs and 6 vacant seats.
The number of ways to arrange them =  ^{12}C_6 = \frac{12!}{6!6!}

The number of ways in which the six teams can be seated in the 30 seats =  ^{12}C_6*6!4! = \frac{12!}{6!6!}*6! = \frac{12!}{6!}

IMO B

In a row of 30 seats, six team of 4 persons each will be seated. The

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