A folk group wants to have one concert on each of the seven consecutive nights...

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A folk group wants to have a concert on each of the seven consecutive nights starting January 1 of next year. One concert is to be held in each of the cities A, B, C, D, and E. Two concerts are to be held in city F, but not on consecutive nights. In how many ways can the group decide on the venues for these seven concerts?

A. 10 x 5!
B. 14 x 5!
C. 15 x 5!
D. 20 x 5!
E. 21 x 5!

OA C

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AAPL wrote:
Fri Mar 06, 2020 4:55 am
GMAT Prep

A folk group wants to have a concert on each of the seven consecutive nights starting January 1 of next year. One concert is to be held in each of the cities A, B, C, D, and E. Two concerts are to be held in city F, but not on consecutive nights. In how many ways can the group decide on the venues for these seven concerts?

A. 10 x 5!
B. 14 x 5!
C. 15 x 5!
D. 20 x 5!
E. 21 x 5!

OA C
Since we want that two concerts must not be scheduled on consecutive nights, let's first place cities except F.

_A_B_C_D_E_

No. of ways, concerts in cities A, B, C, D, and E can be done = 5!

Here a "_" represents places for the concerts in City F. This will ensure that two concerts are not scheduled on consecutive nights.

So, there are 6 "_", and we have to fill two places for City F concerts.

No. of ways 2 slots can be filled out of 6 = 6C2 = (6.5)/(1.2) = 15 ways

Total no. of ways = 15*5!

The correct answer: C

Hope this helps!

-Jay
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AAPL wrote:
Fri Mar 06, 2020 4:55 am
GMAT Prep

A folk group wants to have a concert on each of the seven consecutive nights starting January 1 of next year. One concert is to be held in each of the cities A, B, C, D, and E. Two concerts are to be held in city F, but not on consecutive nights. In how many ways can the group decide on the venues for these seven concerts?

A. 10 x 5!
B. 14 x 5!
C. 15 x 5!
D. 20 x 5!
E. 21 x 5!

OA C
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
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