Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?
6
11
17
72
210
[spoiler]got 7c5 = 21. what on earth to subtract from this ?
oa-17[/spoiler]
p and c
This topic has expert replies
- bblast
- Legendary Member
- Posts: 1079
- Joined: Mon Dec 13, 2010 1:44 am
- Thanked: 118 times
- Followed by:33 members
- GMAT Score:710
Cheers !!
Quant 47-Striving for 50
Verbal 34-Striving for 40
My gmat journey :
https://www.beatthegmat.com/710-bblast-s ... 90735.html
My take on the GMAT RC :
https://www.beatthegmat.com/ways-to-bbla ... 90808.html
How to prepare before your MBA:
https://www.youtube.com/watch?v=upz46D7 ... TWBZF14TKW_
Quant 47-Striving for 50
Verbal 34-Striving for 40
My gmat journey :
https://www.beatthegmat.com/710-bblast-s ... 90735.html
My take on the GMAT RC :
https://www.beatthegmat.com/ways-to-bbla ... 90808.html
How to prepare before your MBA:
https://www.youtube.com/watch?v=upz46D7 ... TWBZF14TKW_
-
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
Hi,
Let shortest to tallest be represented by 1,2,3,4,5,6,7
You have to subtract the case of 4,6 being adjacent.
For this you have select 4,6 but not 5(because if 5 is selected 4 and 6 will not be adjacent)
So, effectively you have to chose the remaining 3 from 1,2,3 and7. This can be done in 4C3 = ways
So, 21-4 =17
Let shortest to tallest be represented by 1,2,3,4,5,6,7
You have to subtract the case of 4,6 being adjacent.
For this you have select 4,6 but not 5(because if 5 is selected 4 and 6 will not be adjacent)
So, effectively you have to chose the remaining 3 from 1,2,3 and7. This can be done in 4C3 = ways
So, 21-4 =17
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
- manpsingh87
- Master | Next Rank: 500 Posts
- Posts: 436
- Joined: Tue Feb 08, 2011 3:07 am
- Thanked: 72 times
- Followed by:6 members
let shortest to tallest be represented by 1,2,3,4,5,6,7bblast wrote:Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?
6
11
17
72
210
[spoiler]got 7c5 = 21. what on earth to subtract from this ?
oa-17[/spoiler]
now 4 and 6 won't be adjacent in following four cases..!!
1) when 4 is not selected.
2) when 6 is not selected.
3) when both 4 and 6 are not selected.
4) when 4,5,6 are selected.
case 1) if 4 is not selected then out of remaining 6 persons 5 persons can be selected in 6C5 ways=6;
these 6 ways would be,
1,2,3,5,7-------------a)
1,2,5,6,7
1,2,3,5,6
1,2,3,6,7
2,3,5,6,7
1,3,5,6,7
case 2) if 6 is not selected then out of remaining 6 persons 5 persons can be selected in 6C5 ways=6;
these 6 ways would be,
1,2,3,5,7------------a)
1,2,3,4,7
1,2,3,4,5
1,2,4,5,7
1,3,4,5,7
2,3,4,5,7
case 3) when both 4 and 6 are not selected then out of remaining 5 persons 5 can be selected in 1 way. i.e. 1,2,3,5,7
case 4) when 4,5,6 are selected, then remaining two persons can be selected from the rest of the 4 persons in 4C2 ways=6;
hence total no. of ways= 6+6+1+6=19,
now if we observe we have counted a) twice in case 1 and 2 and therefore we need to subtract it from the final no. of ways, hence answer should be 19-2=17..!!!
O Excellence... my search for you is on... you can be far.. but not beyond my reach!