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FRUSTRATING PERMUTATIONS AND COMBINATIONS

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FRUSTRATING PERMUTATIONS AND COMBINATIONS

by Sharbel » Tue Apr 01, 2008 4:50 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

CAN ANYBODY SOLVE THIS AND GIVE EXPLANATION PLEASE!!!
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by mandy12 » Tue Apr 01, 2008 6:38 am
IMO the answer is 11.

we have to choose 5 models out of 7 ..so that is 7C5 = 21 ways...

now each of these 21 ways will have exactly 1 way of arranging the models from short to tall .. so total 21 ways to arrange the models ...

the restriction is 4th and 6th tallest model should not be adjacent ...this means all the cases where 4 and 6 are selected but 5 is not selected ......

All the cases where 4 and 6 are selected = 6C4 = 15 ...

All the cases where 4 and 5 and 6 are selected = 5C3 = 10..

so the cases where 4 and 6 are selected but not 5 is = 15 - 10 = 5...

so the total possible cases are ....where both or either of 4 and 6 are missing + where 5 is present with 4 and 6 = (21-15)+5 = 11
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Sorry that's not it

by Sharbel » Tue Apr 01, 2008 11:01 am
i already have the right answer it is 17 but i have no idea how they calculated it.

and regarding the 4 and 6 part i did not understand your logic quite well, can you please explain more.

thank you for the reply anyway
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by mandy12 » Tue Apr 01, 2008 1:47 pm
hmmm .. My mistake ...the answer is indeed 17...
2
as before total possibilities = 21

Now out of these we have to find the ways when 4 and 6 are not adjacent. 4 and 6 can only be adjacent if both of them are selected but 5 is not selected ..(else 5 will come i between 4 and 6 as the numbers represent the height)

so the possibilities when 4 and 6 both are selected
5C3 = 10 (since 4 and 6 are already selected we have to select 3 more out of the remaining 5)..but this includes the cases when 5 is also selected

the possibilities when all 4, 5 and 6 are selected
4C2 = 6

So the cases when only 4 and 6 are selected and 5 is not present -
= 10 -6 = 4

hence the possible arrangements
21 - 4 = 17
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yuppp i think i got it !!!

by Sharbel » Wed Apr 02, 2008 12:23 am
thanks mandy12 i think i got it this time, i just hope that i don't get a similar question in the test cause i don't think i can solve it under pressure. :D
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Re: yuppp i think i got it !!!

by Stuart@KaplanGMAT » Wed Apr 02, 2008 10:50 am
Sharbel wrote:thanks mandy12 i think i got it this time, i just hope that i don't get a similar question in the test cause i don't think i can solve it under pressure. :D
In that case, let's talk about strategic guessing.

We're selecting 5 out of 7. 7C5 = 21. Since they have to stand in order of height, we don't have to worry about rearrangements.

We also have a restriction about 4 and 6. Restrictions will always lower the number of possible combinations.

So, from a guessing point of view, we know that the answer is 21 - something. (d) and (e) are out for sure.

Do we think that the restiction will have a huge impace? If we do, we should choose (a). If we think it will have minimal impact, we should choose (c). If we think it will have medium impact, we should choose (b).
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by Sharbel » Wed Apr 02, 2008 10:43 pm
thank you Stuart,

I did not consider strategic guessing for this exercise since i was solving it with no time constraint and i really wanted to crack it.

But you have a really clear way of putting things together, again thanks a lot.
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