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25 Jun 2018, 08:32
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
78% (01:13) correct
22%
(01:12)
wrong
based on 305
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Bill is renting DVDs from the video store. He must choose three DVDs from a list of eight new releases and then decide in what order to watch them. How many different schedules of DVDs can he create, if he cannot watch the same movie more than once?
(A) 24
(B) 56
(C) 72
(D) 336
(E) 512
(A) 24
(B) 56
(C) 72
(D) 336
(E) 512
25 Jun 2018, 08:39
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CAMANISHPARMAR
It is permutation question.. 8P3=8!/5!=8*7*6=56*6=336
Or you can work on combinations..
First find the possible combinations..
8C5=8!/(5!3!)=8*7*6/6=8*7=56
But these 3 can be arranged in 3! Ways
So total ways =56*3!=336
D
25 Jun 2018, 09:43
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Total no of DVDs = 8
No of DVDs Bill has to select at a time = 3
He can selecting 3 from 8 in \(8c3\) ways.
Selected 3 DVDs can be seen in 3! ways.
So total no of ways = \(8c3\) * 3! = (\(\frac{8!}{3!*5!}\)) * 3! = 8*7*6 = 336
No of DVDs Bill has to select at a time = 3
He can selecting 3 from 8 in \(8c3\) ways.
Selected 3 DVDs can be seen in 3! ways.
So total no of ways = \(8c3\) * 3! = (\(\frac{8!}{3!*5!}\)) * 3! = 8*7*6 = 336
