There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different

A

There are 5 Rock songs, 6 Pop songs, and 3 Jazz: R=5, P=6, J=3
N - the number of combination with at least one Rock song and one Pop song

1.Total number of combination: Nt=2^(5+6+3)=2^14
2. The number of combination without Rock songs: Nr=2^(6+3)=2^9
3. The number of combination without Pop songs: Nr=2^(5+3)=2^8
4. The number of combination without Pop and Rock songs: Nr=2^3
5. N=2^14-2^9-2^8+2^3=16384-512-256+8=15624

https://www.gmatclub.com/forum/t57169 - a similar approach.

Do we not need to know how many songs constitute an 'album' for this question ?

And walker, can you explain the logic between 2^x for total number of combinations ?

1. we have n songs: S={1,2,3,4....,n}
2. each song may be included or not be included in a list. ( two possibilities)
3. our list of songs we can image like a={1,0,0,1,1,0,1,0,....1} - where 1 - in the list, 1 - out of the list.
4. How many lists we can compose? N=2*2*2......2=2^n

can somebody please explain why ''4. The number of combination without Pop and Rock songs: Nr=2^3'' was not subtracted from the overall combinations.

Bunuel I have some doubt on this question, at my understanding the solution of 15624 albums means that one song makes one album correct, If that is the case doesnt that suppouse to be expelcitly stated in the promt. At my understandng and how I approached this question is that i assumed that each album need to have one rock,one pop and one jazz song, so baisicaly one album consists 3 songs of which at least one rock and pop need to be inside teh album. There is my confusion. could you briefly adress my issue here. Thank a lot

Hi kzivrev,

The question keeps a restriction on having at least 1 rock song and 1 pop song in an album. So, an album can have a minimum of 2 songs; that would be the case when there is 1 rock song ,1 pop song and 0 jazz song in the album.

Since, the question does not place any restriction on having a jazz song in the album, an album can be without a jazz song. Hence, your assumption of having minimum of 3 songs in an album is not valid.

If you observe the solution \(2^3*(2^5 - 1)(2^6 - 1) = 15,624.\). Here 1 case has been subtracted from \(2^5\) and \(2^6\) to eliminate the possibility of having 0 rock song or 0 pop song respectively in an album. There is no such elimination of case for jazz songs as there is no restriction of having 1 jazz song in the album.

Hope its clear!

Regards
Harsh

This way of solving the problem a little bit difficult to comprehend. The solution 2^5 * 2^6 * 2^3 - 2^8 - 2^9 + 2^3 looks easier to understand. However, I still bewilder. Please could you help me to figure out. Why we add 2^3 ( only Jazz album ) instead of subtracting them? Why we aren't subtracting albums that contain only Rock and only Pop songs ?

Hi,
when you are subtracting 2^8, you are looking at the combinations of 5 rock songs and 3 jazz songs and
when you are subtracting 2^9, you are looking at the combinations of 6 Pop songs and 3 jazz songs..
so in both cases the jazz songs are subtracted twice , hence we add 2^3..

If you open up 2^3*(2^5 - 1)(2^6 - 1) it will be 2^5 * 2^6 * 2^3 - 2^8 - 2^9 + 2^3

thanks, I think I got it. Could I ask another question about managing albums without songs ( "zero song albums" ) ?
So, do I correctly understand that basically :
2^14 contains one "zero song album"\(R^0 P^0 J^0\)
2^9 contains one "zero song album" too => \(P^0 J^0\)
2^8 contains one "zero song album" too => \(R^0 J^0\)
2^3 contains one "zero song album" too => \(J^0\)

So, at the end, we have \(R^0 P^0 J^0 - P^0 J^0 - R^0 J^0 + J^0 = 0\) => 1("zero song album") - 1("zero song album") - 1("zero song album") + 1("zero song album") = 0 ("zero song album")
To sum up, it is looks like our final solution manages "zero song albums" automatically. That is why we didn't take any additional steps. Is it correct ( sorry if the formula looks a little bit obstruct, hopes you get my concerns ) ?

yes you are correct on your concept..
Now when you have realized this, It will be easier to understand the straight formula..
\(2^3*(2^5 - 1)(2^6 - 1)\)..
\((2^6 - 1)\) is the ways where atleast one POP song is there, so you have removed one case where none were there..
similarily \((2^5 - 1)\) is the ways where atleast one ROCK song is there, so you have removed one case where none were there.
and \(2^3\) remains as it is as it is possible that NO JAZZ song is there..

I understand the method to this problem...but curious as to why (14!/(5!x6!x3!)- 3!) won't work?

Hi
firstly, your formula 14!/(5!6!3!) is for ways in which 14 songs can be arranged where 5 songs of one type and 6 and 3 of other two type..
this is used say when I ask you different words you can form from 'PASSION', which will be 7!/2!..
why should you do this here? AND what does 3! stand for..

Rather it has to be other way " you should tell why this should work?"
May be then someone can find the error

Hi, Bunuel

Sorry, I can not understand the concept at all. please as usual, give the concept first and then give the solution steps.
I always learn from your introduction which may include basics or advanced concepts to save time in test.
Many thanks for your help.

Let's consider in how many ways we can build an album with 5 Rock songs if the albums should contain at least one Rock song. Each of the 5 songs can either be included in the album or not. For example, song #1 can be in the album or can be excluded from the album. Thus each of the 5 songs has 2 options in the album/not in the album. 2*2*2*2*2 = 2^5 = 32 options in total. But those 32 combinations include one combination for which we counted all songs as not included, thus for the album to have at least 1 rock song we should subtract that combination, which gives us 32-1=31 albums with at least 1 rock song.

Hope it helps.

Ok. I got it. I see that this concept is not traditional so when I should know that the solution does not need the traditional formula of Combination or Permutation or such knowledge will be acquired by practice ?
Many thanks Bunuel for caring.

We can use the answer choices to our advantage.

First off, let's ignore the restriction that says "the albums should contain at least one Rock song and one Pop song"
So, we'll find the number of albums possible.
We can do so by taking the task of making albums and break it into STAGES

Let's A = 1st rock song, B = 2nd rock song, . . . . . , M = 2nd Jazz song, and N = 3rd Jazz song.

Stage 1: decide whether to include song A in album
We can choose to have song A or NOT have song A
So, we can complete this stage in 2 ways

Stage 2: decide whether to include song B in album
So, we can complete this stage in 2 ways

Stage 3: decide whether to include song C in album
So, we can complete this stage in 2 ways
.
.
.

Stage 14: decide whether to include song N in album
So, we can complete this stage in 2 ways

By the Fundamental Counting Principle (FCP), we can complete all 14 stages (and thus create an album) in (2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2) ways (= 16,384 ways)

NOTE: One of the 16,384 different albums includes the case in which ZERO songs are selected, which makes no sense.
So, we can subtract 1 from to get a total of 16,383 possible albums (if we IGNORE the restriction)

So, we must now subtract from 16,383 the number of albums that BREAK the restriction.

At this point, we can use the answer choices to our advantage.
First off, we know the correct answer is LESS THAN 16,383, so we can ELIMINATE B

Now consider answer choice C (6,144). This suggests that, among the 16,383 possible albums we created, over 10,000 of them BREAK the rule that says "the albums should contain at least one Rock song and one Pop song"
Does it seem possible that well over half of the 16,383 possible albums BREAK the rule?
No, the rule doesn't seem that restrictive.
So, we can ELIMINATE answer choice C and we can eliminate D and E, since they suggest that almost all of the 16,383 possible albums BREAK the rule

We're left with A

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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