Combination

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Combination

by leonswati » Fri Sep 16, 2011 7:20 pm
A group of 5 friends-Archie, Betty, Jerry, Moose, and Veronica-arrived at the movie theater to see a movie. Because they arrived late, their only seating option consists of 3 middle seats in the front row, an aisle seat in the front row, and an adjoining seat in the third row. If Archie, Jerry, or Moose must sit in the aisle seat while Betty and Veronica refuse to sit next to each other, how many possible seating arrangements are there?
32
36
48
72
120
Source: — Problem Solving |

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by cans » Fri Sep 16, 2011 7:30 pm
A,B,J,M,V.
select 1 to sit for aisle seat: 3C1 =3
case 1: b/v sit in 3rd row: 2C1 (select 1 to send to 3rd row) and arrange the other 3: 3! = 2*6=12
case 2: b/v sit in 1st row only: 2C1 (select one of the other 2 to sit in 3rs row) and as b,v don't sit next to each other, they sit like B-V or V-B -> 2 ways = 2*2 = 4
total = 3*(12+4) = 48
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by GMATGuruNY » Fri Sep 16, 2011 8:50 pm
leonswati wrote:A group of 5 friends-Archie, Betty, Jerry, Moose, and Veronica-arrived at the movie theater to see a movie. Because they arrived late, their only seating option consists of 3 middle seats in the front row, an aisle seat in the front row, and an adjoining seat in the third row. If Archie, Jerry, or Moose must sit in the aisle seat while Betty and Veronica refuse to sit next to each other, how many possible seating arrangements are there?
32
36
48
72
120
Good = Total - Bad.

Total = arrangements with Archie, Jerry or Moose in the aisle seat:
Number of options for the aisle seat = 3. (Archie, Jughead, or Moose)
Number of ways to arrange the 4 other people = 4*3*2*1.
To combine these options, we multiply:
3*4*3*2 = 72.

Bad = arrangements with Archie, Jerry or Moose in the aisle seat BUT with Betty next to Veronica:
Number of options for the aisle seat = 3. (Archie, Jughead, Moose).
Number of options for the third row seat = 2. (Anyone but Betty and Veronica, since in a bad arrangement they sit next to each other.)
Number of options for the middle of the 3 remaining seats = 2. (Must be Betty or Veronica so that they sit next to each other).
Number of ways to arrange the 2 remaining people = 2*1.
To combine these options, we multiply:
3*2*2*2 = 24.

Good arrangements = 72-24 = 48.

The correct answer is C.
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by knight247 » Fri Sep 16, 2011 10:53 pm
We have 3 middle seats, 1 aisle seat and another seat in the 3rd row.
We have friends A B J M V

Total arrangements with no restrictions for Betty and Veronica

In the 1 aisle seat we can only seat either A J or M so 3 ways.
After the above operation there are 4 seats and 4 ppl left. So the 4 ppl can be arranged in the 4 seats in 4!=24 Ways

Total arrangements with no restrictions for Betty and Veronika=24*3=72 Ways

Total arrangements where Betty and Veronica will always be together

Again in the 1 aisle seat we can only seat either A J or M so 3 ways.After performing this operation there are 4 seats and 4 people left.
In the one seat in the 3rd row we can't seat either B or V as they need to be together. So we seat either of the other two so 2 ways After performing this operation there are 3 ppl including B V and another person.Also we have 3 consecutive seats remaining. So we have B V X. We consider BV as one unit so BV X which can be arranged in 2!=2 ways and within the BV unit we have either BV or VB so 2 ways

Total arrangements where Betty and Veronica will always be together=3*2*2*2=24 Ways

72-24=48 Hence C