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30 Dec 2009, 21:43
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
65% (01:41) correct
35%
(01:55)
wrong
based on 1220
sessions
History
Date
Time
Result
Not Attempted Yet
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?
(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!
(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!
31 Dec 2009, 19:53
Kudos
Bookmarks
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?
(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!
There are 3 men and 4 women, we want to calculate the seating arrangements if three men do not sit together, like MMM.
Let's calculate the # of arrangements when they SIT together and subtract from total # of arrangements of these 7 persons without restriction. Thus we'll get the # of arrangements asked in the question.
1. Total # of arrangements of 7 is 7!.
2. # of arrangements when 3 men are seated together, like MMM;
Among themselves these 3 men can sit in 3! # of ways,
Now consider these 3 men as one unit like this {MMM}. We'll have total of 5 units: {MMM}{W}{W}{W}{W}. The # of arrangements of these 5 units is 5!.
Hence total # of arrangements when 3 men sit together is: 3!5!.
# of arrangements when 3 men do not sit together would be: 7!-3!5!.
Answer: C.
Hope it's clear.
(A) 7! – 2!3!2!
(B) 7! – 4!3!
(C) 7! – 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!
There are 3 men and 4 women, we want to calculate the seating arrangements if three men do not sit together, like MMM.
Let's calculate the # of arrangements when they SIT together and subtract from total # of arrangements of these 7 persons without restriction. Thus we'll get the # of arrangements asked in the question.
1. Total # of arrangements of 7 is 7!.
2. # of arrangements when 3 men are seated together, like MMM;
Among themselves these 3 men can sit in 3! # of ways,
Now consider these 3 men as one unit like this {MMM}. We'll have total of 5 units: {MMM}{W}{W}{W}{W}. The # of arrangements of these 5 units is 5!.
Hence total # of arrangements when 3 men sit together is: 3!5!.
# of arrangements when 3 men do not sit together would be: 7!-3!5!.
Answer: C.
Hope it's clear.
31 Dec 2009, 00:31
Kudos
Bookmarks
IMO C
7 people can be seated in 7! ways
take 3 men as one unit ----> tot 5 people can be seated in 5 ways *(no. of ways in which 4 women can be seated amng themselves ) * ( no. of ways in which 3 men cen be seated amng themselves)=5*4!*3!=5!*3!
tot no. of ways in which 3 men are not seated in adjacent seats=tot arrangements - 5!*3!=7!-5!*3!
7 people can be seated in 7! ways
take 3 men as one unit ----> tot 5 people can be seated in 5 ways *(no. of ways in which 4 women can be seated amng themselves ) * ( no. of ways in which 3 men cen be seated amng themselves)=5*4!*3!=5!*3!
tot no. of ways in which 3 men are not seated in adjacent seats=tot arrangements - 5!*3!=7!-5!*3!
