A folk group wants to have one concert on each of the seven consecutiv

7 days - 7 concerts so can be done in 7!/2! Ways
Two consecutive concerts in city F. That can be done in 6! ways

So total ways in non consecutive days is 7!/2-6!= 7/2*6!-6!=6!(5/2)= 6*5!*5/2=15*5!. IMO C

Posted from my mobile device

Ways to arrange A,B,C,D,E = 5!

number of gaps ( including either side gaps)=6
ways to place F in the gaps = 6C2

Total number of ways for mentioned arrangement =6C2*(5!)
Hence, C

There are 7!/2! = (7 x 6 x 5!)/2 = 21 x 5! ways to have concerts if there are no restrictions. However, we are told that they can’t hold a concert in city F on consecutive nights. On the other hand, let’s see how many ways there are if they do hold the concert in city F on consecutive nights. For example, we can have [FF]-A-B-C-D-E. In that case, they can hold the concerts in 6! ways since now FF is considered as one unit. Therefore, there are 21 x 5! - 6! = 5!(21 - 6) = 5!(15) ways to hold the concerts if no two concerts are to be held in city F on consecutive nights.

Answer: C

Hi, ScottTargetTestPrep when considering cases where they do hold the concert in city F on consecutive nights why would we not multiply F by 2? ie 7/2!-6!*2

There is a constraint on city F that the concerts couldn't be held on consecutive nights. This means that there should at least be one city between F's.
_*_*_*_*_*_
In this set-up, if F occupies any of _ and * are occupies by A, B, C, D and E in any possible arrangement then the condition is satisfied.
The permutation between * can happen in 5! ways.
Any 2 of this 6 _ can be selected in 6C2=15 ways
Hence, the total possible number of ways is product of ways of selection and permutation= 15*5!

We want to arrange the letters A, B, C, D, E, F, and F such that the two F's are not adjacent (which would signify two consecutive nights of concerts in city F)

We'll apply the property: # outcomes that satisfy the restriction = (# outcomes that ignore the restriction) - (#outcomes that BREAK the restriction)

In other words: # arrangements with the two F's apart = (# arrangements that ignore the restriction) - (# arrangements with the two F's adjacent)

Let's start with...
# arrangements that ignore the restriction
So, we want to arrange A, B, C, D, E, F, and F in a row.
--------------ASIDE------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
-----------------------------------

With the letters A, B, C, D, E, F, and F, we have:
7 letters in total
2 identical F's
So, the total number of possible arrangements = 7!/2!

# arrangements with the two F's adjacent
To ensure that the two F's are adjacent, let's "glue" them together to create the single object FF.
We now want to arrange the following 6 unique objects: A, B, C, D, E and FF
Since we can arrange n unique objects in n! ways, we can arrange A, B, C, D, E and FF in 6! ways.

So, # arrangements with the two F's apart = 7!/2! - 6!

Looks like we need to express 7!/2! - 6! so that it resembles one of the five answer choices.

To do so let's start by rewriting it as follows: (7)(6)(5!)/2 - (6)(5!)

Since 6/2 = 3, we can rewrite the expression as: (7)(3)(5!) - (6)(5!)

Now factor out 5! to get: (5!)[(7)(3) - 6]

Simplify to get: (5!)(15)

Answer: C

There are 6 cities that need to be arranged in 7 positions because city F is to be placed twice. e.g. A B F C D F E
How many such arrangements are possible? 7!/2! (because two elements are identical)

But arrangements such as F F A B C D E are not allowed. So let's put the two F's together and count them as one group. Now we arrange these in 6! ways. (Note that we don't multiply by 2 here because FF can be placed in only one way because both Fs are identical)
In these ways, both days of F will be together. These are not allowed. So we subtract them out of our possible arrangements.

Acceptable arrangements = 7!/2 - 6! = 5! ( 7*6/2 - 6) = 15*5!

Answer (C)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.