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There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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06 Jul 2004, 20:47
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There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different albums can be formed using the above repertoire if the albums should contain at least one Rock song and one Pop song? A. 15,624 B. 16,384 C. 6,144 D. 384 E. 240
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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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19 Mar 2015, 08:54
Mayanksharma85 wrote: 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. There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different albums can be formed using the above repertoire if the albums should contain at least one Rock song and one Pop song? A. 15,624 B. 16,384 C. 6,144 D. 384 E. 240 There are 3 Jazz songs, each of them can either be included in the album or not, so total of 2 options for each song. Hence, there can be total of 2^3 = 8 different jazz song combination in the album. Notice that those 8 combinations include one combination where none of the jazz songs is included. Similarly, for 5 Rock songs, there are 2^5 combinations. Since 2^5 will also include one case in which there are 0 rock songs, then we should subtract that one case (the albums should contain at least one Rock song) and we'll get 2^5  1. For 6 Rock songs, there are 2^6 combinations: 2^6 will also include one case in which there are 0 pop songs, thus we should subtract that one case (the albums should contain at least one Pop song) and we'll get 2^6  1. Total = 2^3*(2^5  1)(2^6  1) = 15,624. Answer: A. Hope it's clear.
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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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07 Jul 2004, 09:06
If there are 5 rock songs then there are 2^5 ways to make a combination. But there should be atleast 1 Rock song, so the total Rock combination will be 2^5 1
Similar explanation for Pop
But the album can be formed without any Jazz so 2^3
((2^5)1) * ((2^6)1) * 2^3
Ans: 15624




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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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25 Dec 2007, 13:26
There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different albums can be formed using the above repertoire if the albums should contain at least one Rock song and one Pop song?
15624
16384
6144
384
240



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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26 Dec 2007, 04:53
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=2^142^92^8+2^3=16384512256+8=15624
http://www.gmatclub.com/forum/t57169  a similar approach.



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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26 Dec 2007, 17:59
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 ?



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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26 Dec 2007, 23:59
pmenon wrote: 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



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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21 Mar 2008, 00:57
walker, thanks a lot for perfect explanation!



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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21 Mar 2008, 01:00
It is a great example, in which use of nCk and nPk formulas are inappropriate.
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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many
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19 Mar 2015, 08:41
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.



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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19 Mar 2015, 10:45
it is, thanks Bunuel.



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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10 Apr 2015, 07:38
Bunuel, just a small correction, 3rd paragraph of your answer shouldnt it be pop songs ("for 6 rock songs") instead of rock ?



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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05 May 2015, 10:29
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



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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08 May 2015, 00:59
kzivrev wrote: 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
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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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22 Mar 2016, 06:49
Bunuel wrote: Mayanksharma85 wrote: 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. There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different albums can be formed using the above repertoire if the albums should contain at least one Rock song and one Pop song? A. 15,624 B. 16,384 C. 6,144 D. 384 E. 240 There are 3 Jazz songs, each of them can either be included in the album or not, so total of 2 options for each song. Hence, there can be total of 2^3 = 8 different jazz song combination in the album. Notice that those 8 combinations include one combination where none of the jazz songs is included. Similarly, for 5 Rock songs, there are 2^5 combinations. Since 2^5 will also include one case in which there are 0 rock songs, then we should subtract that one case (the albums should contain at least one Rock song) and we'll get 2^5  1. For 6 Rock songs, there are 2^6 combinations: 2^6 will also include one case in which there are 0 pop songs, thus we should subtract that one case (the albums should contain at least one Pop song) and we'll get 2^6  1. Total = 2^3*(2^5  1)(2^6  1) = 15,624. Answer: A. Hope it's clear. 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 ?
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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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22 Mar 2016, 07:03
leeto wrote: Bunuel wrote: Mayanksharma85 wrote: 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. There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different albums can be formed using the above repertoire if the albums should contain at least one Rock song and one Pop song? A. 15,624 B. 16,384 C. 6,144 D. 384 E. 240 There are 3 Jazz songs, each of them can either be included in the album or not, so total of 2 options for each song. Hence, there can be total of 2^3 = 8 different jazz song combination in the album. Notice that those 8 combinations include one combination where none of the jazz songs is included. Similarly, for 5 Rock songs, there are 2^5 combinations. Since 2^5 will also include one case in which there are 0 rock songs, then we should subtract that one case (the albums should contain at least one Rock song) and we'll get 2^5  1. For 6 Rock songs, there are 2^6 combinations: 2^6 will also include one case in which there are 0 pop songs, thus we should subtract that one case (the albums should contain at least one Pop song) and we'll get 2^6  1. Total = 2^3*(2^5  1)(2^6  1) = 15,624. Answer: A. Hope it's clear. 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
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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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25 Mar 2016, 04:08
chetan2u wrote: 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 ) ?
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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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25 Mar 2016, 04:32
leeto wrote: chetan2u wrote: 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..
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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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16 Apr 2016, 06:22
Bunuel wrote: Mayanksharma85 wrote: 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. There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different albums can be formed using the above repertoire if the albums should contain at least one Rock song and one Pop song? A. 15,624 B. 16,384 C. 6,144 D. 384 E. 240 There are 3 Jazz songs, each of them can either be included in the album or not, so total of 2 options for each song. Hence, there can be total of 2^3 = 8 different jazz song combination in the album. Notice that those 8 combinations include one combination where none of the jazz songs is included. Similarly, for 5 Rock songs, there are 2^5 combinations. Since 2^5 will also include one case in which there are 0 rock songs, then we should subtract that one case (the albums should contain at least one Rock song) and we'll get 2^5  1. For 6 Rock songs, there are 2^6 combinations: 2^6 will also include one case in which there are 0 pop songs, thus we should subtract that one case (the albums should contain at least one Pop song) and we'll get 2^6  1. Total = 2^3*(2^5  1)(2^6  1) = 15,624. Answer: A. Hope it's clear. I understand the method to this problem...but curious as to why (14!/(5!x6!x3!) 3!) won't work?



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Re: There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different
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16 Apr 2016, 06:37
Avinashs87 wrote: Bunuel wrote: Mayanksharma85 wrote: 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. There are 5 Rock songs, 6 Pop songs, and 3 Jazz. How many different albums can be formed using the above repertoire if the albums should contain at least one Rock song and one Pop song? A. 15,624 B. 16,384 C. 6,144 D. 384 E. 240 There are 3 Jazz songs, each of them can either be included in the album or not, so total of 2 options for each song. Hence, there can be total of 2^3 = 8 different jazz song combination in the album. Notice that those 8 combinations include one combination where none of the jazz songs is included. Similarly, for 5 Rock songs, there are 2^5 combinations. Since 2^5 will also include one case in which there are 0 rock songs, then we should subtract that one case (the albums should contain at least one Rock song) and we'll get 2^5  1. For 6 Rock songs, there are 2^6 combinations: 2^6 will also include one case in which there are 0 pop songs, thus we should subtract that one case (the albums should contain at least one Pop song) and we'll get 2^6  1. Total = 2^3*(2^5  1)(2^6  1) = 15,624. Answer: A. Hope it's clear. 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
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