arrange with restriction ?????

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

AIM GMAT: Legendary Member; Posts: 857; Joined: Wed Aug 25, 2010 1:36 am; Thanked: 56 times; Followed by:15 members

arrange with restriction ?????

by AIM GMAT » Sun Mar 20, 2011 1:44 am

Tanya bought 6 glasses - white, red, black, grey, blue and yellow - and
would like to display 3 of them on the shelf next to each other. If she decides that a red
and a blue glass cannot be displayed at the same time, in how many different ways can Tanya
arrange the glasses?

Thanks & Regards,
AIM GMAT

Quote

Source: Beat The GMAT — Problem Solving | Sun Mar 20, 2011 1:44 am

srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 am; Location: Seattle, WA; Thanked: 86 times; Followed by:2 members

by srcc25anu » Sun Mar 20, 2011 2:16 am

total ways to arrange 6 glasses taken 3 at a time = 6P3 = 120
assuming we select both red and blue glasses, ways to arrange = 3.2.4C1 = 24 ways
ways to arrange 6 glasses with restriction that red and blue glass are not taken together = 120 - 24 = 96

what's the OA?

Quote

AIM GMAT: Legendary Member; Posts: 857; Joined: Wed Aug 25, 2010 1:36 am; Thanked: 56 times; Followed by:15 members

by AIM GMAT » Sun Mar 20, 2011 2:19 am

total ways to arrange 6 glasses taken 3 at a time = 6P3 = 120
assuming we select both red and blue glasses, ways to arrange = 3.2.4C1 = 24 waysways to arrange 6 glasses with restriction that red and blue glass are not taken together = 120 - 24 = 96

OA is 96 . I dint get the step that's highlighted . Can u explain in detail ?

Thanks & Regards,
AIM GMAT

Quote

srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 am; Location: Seattle, WA; Thanked: 86 times; Followed by:2 members

by srcc25anu » Sun Mar 20, 2011 2:24 am

we are taking the scenario of (red , blue and one other color)
first a red glass can be arranged in 3 ways
a blue glass can then be arranged in 2 ways
For the remaining third place, we can select 1 out of 4 colors in 4C1 ways and its position can be just be the one remaining after we have placed red and blue in 2 out of 3 places
hence taking all the ways its 3*2*4 ways = 24

and since these are unfavorable. cases, we subtract them from total 120 to get 96 desired outcomes

hope thats clear

AIM GMAT wrote:total ways to arrange 6 glasses taken 3 at a time = 6P3 = 120
assuming we select both red and blue glasses, ways to arrange = 3.2.4C1 = 24 waysways to arrange 6 glasses with restriction that red and blue glass are not taken together = 120 - 24 = 96

OA is 96 . I dint get the step that's highlighted . Can u explain in detail ?