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by bblast » Mon Jun 20, 2011 8:41 am
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]
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by Frankenstein » Mon Jun 20, 2011 8:49 am
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
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by manpsingh87 » Mon Jun 20, 2011 10:18 am
bblast 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]
let shortest to tallest be represented by 1,2,3,4,5,6,7

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..!!!
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