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Craig invited four friends to watch a TV show. He arranged 5
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16 May 2019, 00:14
16 May 2019, 00:14
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Question Stats:
70% (01:10) correct
29% (00:38) wrong based on 17 sessions
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Craig invited four friends to watch a TV show. He arranged 5 seats in a row. The number of ways he and his four friends can sit in the row is n. In how many of these ways can Craig sit in the middle?
(A) \(n\)
(B) \(\frac{n}{2}\)
(C) \(\frac{n}{3}\)
(D) \(\frac{n}{4}\)
(E) \(\frac{n}{5}\)
(A) \(n\)
(B) \(\frac{n}{2}\)
(C) \(\frac{n}{3}\)
(D) \(\frac{n}{4}\)
(E) \(\frac{n}{5}\)
Re: Craig invited four friends to watch a TV show. He arranged 5
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22 Aug 2019, 18:17
22 Aug 2019, 18:17
Carcass wrote:
Craig invited four friends to watch a TV show. He arranged 5 seats in a row. The number of ways he and his four friends can sit in the row is n. In how many of these ways can Craig sit in the middle?
(A) \(n\)
(B) \(\frac{n}{2}\)
(C) \(\frac{n}{3}\)
(D) \(\frac{n}{4}\)
(E) \(\frac{n}{5}\)
(A) \(n\)
(B) \(\frac{n}{2}\)
(C) \(\frac{n}{3}\)
(D) \(\frac{n}{4}\)
(E) \(\frac{n}{5}\)
5! is n and is how many ways they can all sit and 4! is how many combinations there are with him in the middle.
Therefor 4! = (x)5!
4!/5! = 1/5 = x
substitute x =1/5 and 5! = n into the equation to see that n/5 indeed the answer (e).
gmatclubot
Re: Craig invited four friends to watch a TV show. He arranged 5 [#permalink]
22 Aug 2019, 18:17
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