I have a big question about permutation here. the first question, I can understand the way how to solve. but I used the same way to solve the 2nd question, it was wrong, please help me here!!! why I can't use the same way to solve the 2nd one??
question 1: 8 cars (3red, 3blue, 2 yellow)are to be parked in a line, how many unique lines can be formed if the yellow cars must not be together? assume that cars of each colors are identical.
8!-2!*7!=18
question 2: how many ways can 8 books, each covering a different subject, be arranged on a shelf such that books on biology, history, or programming are never together?
8!-3!*5!= this is wrong, but I don't get it!
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permutation help needed here!!!
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Hi svbla,
When posting questions, you should make sure to post one per thread (this helps keep each conversation focused on one prompt). Also, you should make sure to include the full question WITH the 5 answer choices (sometimes the answer choices provide a big hint as to how you can go about answering the question).
GMAT assassins aren't born, they're made,
Rich
When posting questions, you should make sure to post one per thread (this helps keep each conversation focused on one prompt). Also, you should make sure to include the full question WITH the 5 answer choices (sometimes the answer choices provide a big hint as to how you can go about answering the question).
GMAT assassins aren't born, they're made,
Rich
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Matt@VeritasPrep
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Q1:
We've got
8! / (3! * 3! * 2!) total arrangements
To find the ones in which the yellow cars are together, treat them as one car, so we have 7 total, with 3 red and 3 blue.
7! / (3! * 3!) invalid arrangements
So our total is 8!/(3!*3!*2!) - 7!/(3!*3!), or 420.
We've got
8! / (3! * 3! * 2!) total arrangements
To find the ones in which the yellow cars are together, treat them as one car, so we have 7 total, with 3 red and 3 blue.
7! / (3! * 3!) invalid arrangements
So our total is 8!/(3!*3!*2!) - 7!/(3!*3!), or 420.
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Matt@VeritasPrep
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Q2:
This is a tough one, partly because it's ambiguous: does it mean that all three can't be together in a row, or that any two can't be together? I'm going to assume it means that no two can be adjacent.
We've got 8! arrangements with no restrictions, so we'll start there.
If the bio and history books are together, we'll treat them as one item. Then we have 7! * 2 ways of arranging our 7 books: 7 because we've counting bio+history as one book, and *2 because even though the bio and history count as one book, they can be arranged in two ways next to each other in their space (bio on the left or history on the left).
Bio+programming and history+programming would be the same: 7! * 2 arrangements each.
Now we have to add back bio+history+programming all together, since we've subtracted these arrangements from each of our three pairs above. Here we have six books (since we're treating these three books as one), and we can arrange them in their space in six ways. That gives us 6! * 6.
In all, then, we have
8! - 7!*2 - 7!*2 - 7!*2 + 6!*6, or 14,400.
This is a tough one, partly because it's ambiguous: does it mean that all three can't be together in a row, or that any two can't be together? I'm going to assume it means that no two can be adjacent.
We've got 8! arrangements with no restrictions, so we'll start there.
If the bio and history books are together, we'll treat them as one item. Then we have 7! * 2 ways of arranging our 7 books: 7 because we've counting bio+history as one book, and *2 because even though the bio and history count as one book, they can be arranged in two ways next to each other in their space (bio on the left or history on the left).
Bio+programming and history+programming would be the same: 7! * 2 arrangements each.
Now we have to add back bio+history+programming all together, since we've subtracted these arrangements from each of our three pairs above. Here we have six books (since we're treating these three books as one), and we can arrange them in their space in six ways. That gives us 6! * 6.
In all, then, we have
8! - 7!*2 - 7!*2 - 7!*2 + 6!*6, or 14,400.
