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17 Mar 2021, 05:30
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (02:54) correct
59%
(02:41)
wrong
based on 58
sessions
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There are 3 different books on Mathematics, 4 different books on Geography and 6 different books on Microbiology, which are to be arranged on a shelf. Let A be the event that no two books on Microbiology are together and B be the event that all the books on the same subject are always together. What is the ratio of the probability of the event A occurring to that of the probability of event B ?
(A) 3/490
(B) 75/221
(C) 81/65
(D) 30/7
(E) 490/3
(A) 3/490
(B) 75/221
(C) 81/65
(D) 30/7
(E) 490/3
24 Mar 2021, 09:41
Kudos
Bookmarks
Bunuel
Probability of A: Arrange the first 7 books on Mathematics and Geography first, there are 7! arrangements for that. Then we slot in the 6 different books on Microbiology. The 7 books create 8 slots to slot in Microbiology books, the two ends and a slot between neighboring books. The first Microbiology book has 8 slots, the next one has 7 slots, etc. So 8P6 slots for Microbiology books. In total \(\frac{7!*8!}{2}\) arrangements. Divide by 13! for probability.
Probability of B: There are 3 subjects, so 3! = 6 arrangements of subjects. Within each subject we can randomly scramble. Thus in total 6*3!*4!*6! arrangements. Divide by 13! for probability.
Finally we need to find the ratio of A over B, as we can see the 13! cancels. Leaving us with \(\frac{7!*8! }{ 2*6*3!*4!*6!} = \frac{7*6*5*8*7}{2*6*3!} = \frac{7*5*2*7 }{ 3}= \frac{490}{3}\).
Ans: E
29 Mar 2021, 11:16
Kudos
Bookmarks
Bunuel
Solution:
Let M denote a microbiology book and X denote a mathematics or a geography book. If no two microbiology books are together, then one of the following must be true:
1) The arrangement contains no X-X or X-X-X:
There is only one arrangement satisfying this condition, namely X-M-X-M-X-M-X-M-X-M-X-M-X.
2) The arrangement contains one X-X:
One arrangement satisfying this condition is M-X-X-M-X-M-X-M-X-M-X-M-X. Moving the X-X pair around, we see that there are six arrangements in this configuration.
Another arrangement with one X-X is X-X-M-X-M-X-M-X-M-X-M-X-M. Using a similar argument as above, there are six additional arrangements.
3) The arrangement contains two X-X:
One arrangement satisfying this condition is M-X-X-M-X-X-M-X-M-X-M-X-M. Since there are five spaces between M’s, there are 5C2 = 5!/(2!*3!) = 10 arrangements in this category.
4) The arrangement contains one X-X-X:
One arrangement satisfying this condition is M-X-X-X-M-X-M-X-M-X-M-X-M. Moving the X-X-X around, we see that there are five such arrangements.
In total, there are 1 + 6 + 6 + 10 + 5 = 28 possible configurations. For each configuration, the microbiology books can be arranged in 6! ways, and the math and geography books can be arranged in 7! ways. Thus, there are 28 x 6! x 7! arrangements where no two microbiology books are together.
Since there are 13 books in total, there are 13! arrangements in total. Thus, probability of event A is (28 x 6! x 7!)/13!.
For the probability of event B, the three subjects can be arranged in 3! ways. The mathematics, geography and microbiology books can be arranged in 3!, 4! and 6! ways, respectively. Thus, the probability of event B is (3! x 3! x 4! x 6!)/13!.
The ratio of the probabilities is [(28 x 6! x 7!)/13!] / [(3! x 3! x 4! x 6!)/13!] = (28 x 6! x 7!) / (3! x 3! x 4! x 6!) = (28 x 7 x 6 x 5) / (3! x 3!) = (28 x 35) / 6 = 490/3.
Answer: E
