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# How many different arrangements to seat four spectators in four of sev

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Math Expert
Joined: 02 Sep 2009
Posts: 58415
How many different arrangements to seat four spectators in four of sev  [#permalink]

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22 Mar 2018, 23:57
00:00

Difficulty:

35% (medium)

Question Stats:

58% (01:31) correct 42% (01:11) wrong based on 51 sessions

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How many different arrangements to seat four spectators in four of seven empty stadium seats that are all in the same row are possible?

(A) 12
(B) 24
(C) 28
(D) 35
(E) 840

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Re: How many different arrangements to seat four spectators in four of sev  [#permalink]

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23 Mar 2018, 00:11
Bunuel wrote:
How many different arrangements to seat four spectators in four of seven empty stadium seats that are all in the same row are possible?

(A) 12
(B) 24
(C) 28
(D) 35
(E) 840

Since there are 7 empty seats and we need to seat four spectators in four of
the empty stadium seats, there are 7 ways of seating the first spectator.
There are 6,5, and 4 ways of seating the other 3 spectators.

Therefore, there are 7*6*5*4 = 42*20 = 840 ways(Option E) of seating the 4 spectators
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Re: How many different arrangements to seat four spectators in four of sev  [#permalink]

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23 Mar 2018, 15:48
Let's say the 4 spectators are A,B,C,D. Can we assume that these arrangements are all different:
A,B,C,D,x,x,x
A,B,C,x,D,x,x
A,B,C,x,x,D,x
A,B,C,x,x,x,D

Assuming yes, my approach was to find the number of ways to arrange 4 ppl in 4 seats and the number of ways to select 3 out of 7 empty seats (order doesnt matter with empty seats). Multiplying the two will get me the answer.
# of ways to arrange 4 ppl: 4! = 24
# of ways to select 3 empty seats: 7C3 = 35
24 * 35 = 840

Re: How many different arrangements to seat four spectators in four of sev   [#permalink] 23 Mar 2018, 15:48
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