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

Question

A folk group wants to have one concert on each of the seven consecutive nights starting January 1 of next year. One concert is to be held in each of 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!

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

CareerGeek · Accepted Answer

Total ways = arranging A, B, C, D, E, F, F = 7!/2!

Calculate the number of ways that 2 shows in city F are done together 
= take FF as a single bock
=  , A, B, C, D, E
= 6!*2!/2!

So, required number of ways = 7!/2! - 6!*2!/2!
= (7*6*5! - 2*6*5!)/2
= (42 - 12)*5!/2
= 15*5!

IMO Option C

IanStewart · Answer

We can arrange the cities A, B, C, D and E in 5! ways. We'd then get some ordering like this:

_ D _ A _ B _ E _ C _

I've inserted above the six slots where we could put in our two "F"s so that the two F's are not together. Since order does not matter, we can choose 2 of those 6 slots in 6*5/2! = 15 ways. So the answer is 15*5!.

This is a very old GMATPrep question, by the way.

Lampard42 · Answer

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

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pa1deep91 · Answer

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

ScottTargetTestPrep · Answer

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  -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

rajdev98 · Answer

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!

BrentGMATPrepNow · Answer

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

KarishmaB · Answer

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)

steffe440 · Answer

concerts F can never be consecutive hence we have these possible outcomes

_A_B_C_D_E_ so concerts F can be in the gap as long as they are not in the same gap,

Hence what we have is F can take place in 6C2 ways! Or 15 ways. While A B C D and E can be arranged in 5!

Hence we get 15*5!

iamyogi25 · Answer

@bunuel.. why are we not dividing 6! (when we glued FF together) by 2! just liked we did 7! by 2! for the total cases? Thank you

Bunuel · Answer

When FF is glued together, it becomes  one single block . So the six items are:

{FF}, {A}, {B}, {C}, {D}, {E}

These are 6   distinct   items, so the number of arrangements is 6!.

We divide 7! by 2! in the total count because there the two F's are separate identical items, and swapping them does not create a new arrangement.

kevincan · Answer

Wonderful question !

egmat · Answer

Hi iamyogi25,

Good instinct to look for consistency, but the two situations aren't the same, so the  2!  belongs in one and not the other. Let me pin down exactly what that  2!  is  fixing .

What the 2! actually corrects.  In the total count, we start by pretending the two F's are distinct - call them F1 and F2 - and arrange  7  distinct items in  7!  ways. But since the two F's are really identical, every real arrangement got counted  twice  (once as ...F1...F2... and once as ...F2...F1...). Dividing by  2!  removes that double-count. So the  2!  is there  only to undo a doubling that distinct-labeling created.

Why the glued block has nothing to undo.  When you glue the F's into one block   and arrange the  6  objects A, B, C, D, E,  , you never labeled the F's as distinct in the first place. The block is just "FF" - there is only  one  way to sit two identical F's inside it. So  6!  already counts each arrangement exactly once. There's no doubling to cancel, so there's no  2!  to divide by. Dividing again would wrongly cut your count in half.

Another way to see it: if the F's  were  distinct, the glued block could be F1F2 or F2F1, giving  6! × 2! . Then dividing by  2!  for "identical" brings you right back to  6! . Either way you land on  6! .

Quick check with tiny numbers  - arrange A, F, F:
- All arrangements:  3!/2! = 3  - they are  FFA, FAF, AFF .
- F's adjacent (glue them): arrange A and   -  2  ways -  FFA, AFF . Count them in the list: yes, exactly  2 .

If you had divided that by another  2! , you'd get  1  - but you can clearly see there are  2 . That's the proof the extra division doesn't belong.

So:  7!/2! - 6! = 15 × 5!

Answer: C

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