gmatusa2010 wrote:wow. can you expand a little bit here?
anshumishra wrote:gmatusa2010 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 cannot be adjacent, how many different
arrangements of five models are possible?
Choose 4 and 6, we must also choose 5 who will stand between them =3C3*4C2=4C2=6
Choose either 4 or 6 = 2*1*5C4=10
Choose neither 4 nor 6 = 5C5=1
So, total no. of arrangements = 17.
Sure.
There are only 3 possible cases. We choose
both 4 and 6,
Either 4 or 6 , Or
Neither 4 nor 6
Choose 4 and 6, we must also choose 5 who will stand between them =
3C3*4C2=4C2=6 {3C3 -> No. of ways of choosing 4,5 and 6 out of 4,5 and 6. 4C2-> No. of ways of selecting the remaining 2 out of 4 left}
Choose either 4 or 6 = 2*1*5C4=10 {No. of ways to select either 4 or 6 is 2*1 or just say 2, 5C4 -> choose the remaining 4 from the other 5 left}
Choose neither 4 nor 6 = 5C5=1 {If you choose neither 4 nor 6, you are left with remaining 5 models. So number of ways of selecting 5 models out of 5 = 5C5 or just say 1}
So, total no. of arrangements = 17.
Let me know if you have any doubts.[/i]
Thanks
Anshu
(Every mistake is a lesson learned )
