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07 Feb 2008, 13:08
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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:
65%
(hard)
Question Stats:
56% (01:59) correct
44%
(01:51)
wrong
based on 469
sessions
History
Date
Time
Result
Not Attempted Yet
Jake, Lena, Fred, John and Inna need to drive home from a corporate reception in an SUV that can seat 7 people. If only Inna or Jake can drive, how many seat allocations are possible?
A. 30
B. 42
C. 120
D. 360
E. 720
A. 30
B. 42
C. 120
D. 360
E. 720
07 Feb 2008, 13:15
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Bookmarks
E
[X][YYYYYY]
X: Inna or Jake: \(P^2_1\)
Y: Lena , Fred, John and (Jake or Inna): \(P^6_4\)
\(N=P^2_1*P^6_4=2*\frac{6!}{2!}=6*5*4*3*2=720\)
[X][YYYYYY]
X: Inna or Jake: \(P^2_1\)
Y: Lena , Fred, John and (Jake or Inna): \(P^6_4\)
\(N=P^2_1*P^6_4=2*\frac{6!}{2!}=6*5*4*3*2=720\)
nikunjhd
Joined: 16 Apr 2007
Last visit: 31 May 2014
Posts: 7
Location: United States
Schools: Haas EWMBA '17 (A)
07 Feb 2008, 17:34
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E.720
5 people ,7 seats, only 2 can drive
if inna is in one seat remaining, 4 people can be arranged in 4p4 i.e 4! =24 ways
since any one can drive total ways = 24 *2= 48 ways
4 seats can be chosen out of remaining 6 seats in 6c4 ways = 6!/(4!*2!)= 15 ways
48*15=720
5 people ,7 seats, only 2 can drive
if inna is in one seat remaining, 4 people can be arranged in 4p4 i.e 4! =24 ways
since any one can drive total ways = 24 *2= 48 ways
4 seats can be chosen out of remaining 6 seats in 6c4 ways = 6!/(4!*2!)= 15 ways
48*15=720
