BTGmoderatorLU wrote: ↑

Tue May 26, 2020 6:27 am

Source: Magoosh

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. (12!)/(6!)
C. 12C6
D. 30C12
E. 30C6*6

The OA is B

Since all members of a team are to sit together, let's club 4 members of each team. Say the teams are T1, T2, T3, T4, T5, and T6. So, we have 30 – 6*4 = 6 empty seats and 6-sets of the occupied seat, with each set has 4 seats.

So, the situation is T1, T2, T3, T4, T5, T6, E, E, E, E, E, E; where E = empty seat

So, the no. of ways 12 sets can be occupied out of 12 sets = 12!/6!; I divided 12! by 6! since 6 Es are identical.

The correct answer: B

Hope this helps!

-Jay
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BTGmoderatorLU wrote: ↑

Tue May 26, 2020 6:27 am

Source: Magoosh

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. (12!)/(6!)
C. 12C6
D. 30C12
E. 30C6*6

The OA is B

Solution:

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