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The New York Classical Group is designing the liner notes for an upcom

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The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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02 Jul 2015, 02:21
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Difficulty:

15% (low)

Question Stats:

74% (00:54) correct 26% (01:03) wrong based on 195 sessions

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The New York Classical Group is designing the liner notes for an upcoming CD release. There are 10 soloists featured on the album, but the liner notes are only 5 pages long, and therefore only have room for 5 of the soloists. The soloists are fighting over which of them will appear in the liner notes, though not about which page they appear on. How many different combinations of soloists can appear in the liner notes?

A. 5!
B. 10!/(5!5!)
C. 10!/5!
D. 10!
E. 10!*5!

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Re: The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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02 Jul 2015, 02:26
1
There are 10 soloists ------------- (1)
Room available is only for 5 soloists -------------- (2)

There is no issue for sequencing, i.e selected 5 soloists say for eg : A,B,C,D,E are same as C,E,D,B,A

Therefore from (1) & (2),
Number of ways of selecting 5 different soloists from 10 is 10C5
= 10!/(5!*5!)
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The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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02 Jul 2015, 22:32
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1
Bunuel wrote:
The New York Classical Group is designing the liner notes for an upcoming CD release. There are 10 soloists featured on the album, but the liner notes are only 5 pages long, and therefore only have room for 5 of the soloists. The soloists are fighting over which of them will appear in the liner notes, though not about which page they appear on. How many different combinations of soloists can appear in the liner notes?

A. 5!
B. 10!/(5!5!)
C. 10!/5!
D. 10!
E. 10!*5!

Kudos for a correct solution.

Answer: 1st page can be filled in 10 ways, 2nd page in 9 ways,....5th page in 6 ways.
therefore, total combination = 10*9*8*7*6 = (10!)/5!
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Re: The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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02 Jul 2015, 23:11
5 pages to be filled by 10 soloists
1st page can be filled by in 10 ways
2nd page - 9 ways
3rd page - 8 ways
4th page - 7 ways
5th page - 6 ways

total = 10*9*8*7*6 = 10!/5! = C?
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The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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03 Jul 2015, 00:54
Bunuel wrote:
The New York Classical Group is designing the liner notes for an upcoming CD release. There are 10 soloists featured on the album, but the liner notes are only 5 pages long, and therefore only have room for 5 of the soloists. The soloists are fighting over which of them will appear in the liner notes, though not about which page they appear on (i.e. NO ARRANGEMENT). How many different combinations of soloists can appear in the liner notes?

A. 5!
B. 10!/(5!5!)
C. 10!/5!
D. 10!
E. 10!*5!

Kudos for a correct solution.

Given:
Total Soloists to choose from = 10
Soloist Required = 5

Question : How many different combinations of soloists can appear in the liner notes?

The language of the question clearly mentions that only the combinations need to be calculated

i.e. Arrangement of the Soloists does NOT matter (Refer the Highlighted part of the question)

Method-1:

Total Ways to Choose 5 out of 10 soloists = 10C5 = 10! / (5!*5!)

Method-2:

No. of ways to select and arrange 5 soloist on 5 Page = 10*9*8*7*6

Since the arrangement of 5 selected Notes (which can happen in 5! ways) doesn't NOT matter,

Therefore total ways to pick 5 out of 10 soloists = 10*9*8*7*6 / 5! = 10*9*8*7*6 *5! / (5!*5!) = 10! / (5!*5!)

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Re: The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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03 Jul 2015, 11:05
simply .. 10c5 = 10!/5!5!
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Re: The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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04 Jul 2015, 10:10
1
Bunuel wrote:
The New York Classical Group is designing the liner notes for an upcoming CD release. There are 10 soloists featured on the album, but the liner notes are only 5 pages long, and therefore only have room for 5 of the soloists. The soloists are fighting over which of them will appear in the liner notes, though not about which page they appear on. How many different combinations of soloists can appear in the liner notes?

A. 5!
B. 10!/(5!5!)
C. 10!/5!
D. 10!
E. 10!*5!

Kudos for a correct solution.

No of soloists=10
No of liner notes=5
Therefore, no of combinations=10C5=10!/5!*5!
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Re: The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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04 Jul 2015, 12:15
Option B is correct
No of ways
10C 5= 10!/5!*5!
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Posts: 49206
Re: The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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06 Jul 2015, 06:51
Bunuel wrote:
The New York Classical Group is designing the liner notes for an upcoming CD release. There are 10 soloists featured on the album, but the liner notes are only 5 pages long, and therefore only have room for 5 of the soloists. The soloists are fighting over which of them will appear in the liner notes, though not about which page they appear on. How many different combinations of soloists can appear in the liner notes?

A. 5!
B. 10!/(5!5!)
C. 10!/5!
D. 10!
E. 10!*5!

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

In this problem, the order in which the soloists appear is important. Therefore, the problem can be modeled with anagrams of the “word” 12345NNNNN, in which each number represents the page on which a soloist might appear:

Attachment:

2015-07-06_1750.png [ 12.71 KiB | Viewed 2391 times ]

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Re: The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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11 Jun 2016, 09:31
NickHalden wrote:
There are 10 soloists ------------- (1)
Room available is only for 5 soloists -------------- (2)

There is no issue for sequencing, i.e selected 5 soloists say for eg : A,B,C,D,E are same as C,E,D,B,A

Therefore from (1) & (2),
Number of ways of selecting 5 different soloists from 10 is 10C5
= 10!/(5!*5!)

What if the order did matter? Then would the answer be 10C5 * 5! ???
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Re: The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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11 Jul 2018, 01:24
Bunuel wrote:
The New York Classical Group is designing the liner notes for an upcoming CD release. There are 10 soloists featured on the album, but the liner notes are only 5 pages long, and therefore only have room for 5 of the soloists. The soloists are fighting over which of them will appear in the liner notes, though not about which page they appear on. How many different combinations of soloists can appear in the liner notes?

A. 5!
B. 10!/(5!5!)
C. 10!/5!
D. 10!
E. 10!*5!

Kudos for a correct solution.

there are 10 soloist and only 5 will be chosen. The sequence of choice is not important.

Soloist Number chosen(Yes/No)
1 - y
2 - y
3 - y
4 - y
5 - y
6 - n
7 - n
8 - n
9 - n
10 - n

so there are 10! total arrangements and we need to remove the 5! yes(selected) and 5! No(not selected)
$$\frac{10!}{5!*5!}$$
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Re: The New York Classical Group is designing the liner notes for an upcom  [#permalink]

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17 Jul 2018, 16:41
Bunuel wrote:
The New York Classical Group is designing the liner notes for an upcoming CD release. There are 10 soloists featured on the album, but the liner notes are only 5 pages long, and therefore only have room for 5 of the soloists. The soloists are fighting over which of them will appear in the liner notes, though not about which page they appear on. How many different combinations of soloists can appear in the liner notes?

A. 5!
B. 10!/(5!5!)
C. 10!/5!
D. 10!
E. 10!*5!

Since they are not fighting about which page they will appear on, the order the 5 soloists chosen doesn’t matter. If order doesn’t matter, we use combinations. So the answer is 10C5 = 10!/(5!(10 - 5)!) = 10!/(5!5!).

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Re: The New York Classical Group is designing the liner notes for an upcom &nbs [#permalink] 17 Jul 2018, 16:41
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