There are 5 maths books,4 science books,and 3 literature books.How many different collections a person can make selecting at least one of each kind when all books are different?and when the books in a category are same?
Ans choices: 1)4095,60 2)3255,60 3)4095,55 4)3255,59
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vipulgoyal
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Experts please suggest
i guess here lot of calculations involve for the first part there will be 60 arrangements involved
and we have to calculate all one by one then add up for ex 5c1*4c1*3c1 + 5c2*4c1*3c1 + ...... 5c5*4c4*3c3 the second ans is 60 , at least now we come up with 2 options 1 and 2
i guess here lot of calculations involve for the first part there will be 60 arrangements involved
and we have to calculate all one by one then add up for ex 5c1*4c1*3c1 + 5c2*4c1*3c1 + ...... 5c5*4c4*3c3 the second ans is 60 , at least now we come up with 2 options 1 and 2
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vipulgoyal
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Brent@GMATPrepNow
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Wow - I read the question 3 times, and I'm still not 100% sure what it's asking.
The strange answer choices and the fact that there are only 4 options suggests that this question is not GMAT-worthy.
Cheers,
Brent
The strange answer choices and the fact that there are only 4 options suggests that this question is not GMAT-worthy.
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Matt@VeritasPrep
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Vipul, good question - it's actually a bit simpler than that, but it relies on a bit of set theory that the GMAT doesn't often test ... which, along with the four answer choices, makes me suspect that this is an unofficial question someone wrote to help students prepare for India's CAT.
Let's take the 'literature books' to start. For each of the three books, I have one of two choices: the book is in my collection, or the book is not in my collection. So I have a total of 2*2*2 = 8 ways of making a set of those books. To illustrate, let's call the books A, B, and C:
[no A, no B, no C]
[A, no B, no C]
[no A, B, no C]
[no A, no B, C]
[no A, B, C]
[A, no B, C]
[A, B, no C]
[A, B, C]
But this question stipulates that I have at least one 'literature book' (maybe English As She Is Spoke?
) in my collection, so I really only have SEVEN options - [no A, no B, no C] is out.
From there, we can extend the principle to the science and math books, for which we'll have 2*2*2*2*2-1 = 31 and 2*2*2*2-1 = 15 possible libraries, respectively. This gives us 31 science libraries, 15 math libraries, and 7 literature libraries, and since we must have one library from each group, we have a total of 31 * 15 * 7 = 3255 possible libraries consisting of at least one book from each group. (The trap answer 4095 is cute: this is what you get if you multiply 32 * 16 * 8, then subtract 1, i.e. if you read the question as requiring at least one BOOK, not one of each type.)
The second part of the question is a little easier, but I'll leave that for students to discuss. (It's no fun to just give everything away, right?)
Let's take the 'literature books' to start. For each of the three books, I have one of two choices: the book is in my collection, or the book is not in my collection. So I have a total of 2*2*2 = 8 ways of making a set of those books. To illustrate, let's call the books A, B, and C:
[no A, no B, no C]
[A, no B, no C]
[no A, B, no C]
[no A, no B, C]
[no A, B, C]
[A, no B, C]
[A, B, no C]
[A, B, C]
But this question stipulates that I have at least one 'literature book' (maybe English As She Is Spoke?
) in my collection, so I really only have SEVEN options - [no A, no B, no C] is out. From there, we can extend the principle to the science and math books, for which we'll have 2*2*2*2*2-1 = 31 and 2*2*2*2-1 = 15 possible libraries, respectively. This gives us 31 science libraries, 15 math libraries, and 7 literature libraries, and since we must have one library from each group, we have a total of 31 * 15 * 7 = 3255 possible libraries consisting of at least one book from each group. (The trap answer 4095 is cute: this is what you get if you multiply 32 * 16 * 8, then subtract 1, i.e. if you read the question as requiring at least one BOOK, not one of each type.)
The second part of the question is a little easier, but I'll leave that for students to discuss. (It's no fun to just give everything away, right?)
Selection of 0-n objects from n different objects is given by (2^n)
Selection of 1-n objects from n different objects is (2^n-1).
Therefore we have
Total ways of selecting 1-n books in each category : (2^5-1)* (2^4-1) * (2^3-1)
Answer = 3255..
Second part is east. (5)*(4)*(3)=60
answer is 3255 , 60
Selection of 1-n objects from n different objects is (2^n-1).
Therefore we have
Total ways of selecting 1-n books in each category : (2^5-1)* (2^4-1) * (2^3-1)
Answer = 3255..
Second part is east. (5)*(4)*(3)=60
answer is 3255 , 60
