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probability problem with condition

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by r_walid » Sun Nov 15, 2009 12:32 pm
out of 7 models, 5 will be selected for a photo. if the 5 models are to stand in a line from shortest to tallest and if all are of different heights, and if the fourth and sixth tallest models cannot sit side by side, how many different arrangements of 5 models are possible?

6
11
17
72
210

Please explain with DETAIL your reasoning.
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by palvarez » Mon Nov 16, 2009 11:21 am
7c5 = 21 ways without any restrictions.

4 and 6 can't sit side by side, meaning that whenever 4 and 6 are present, we need 5.

Find ways where 4 and 6 are present, but not 5: 4c3 ways = 4

21 - 4 = 17.
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by gmat620 » Wed Nov 18, 2009 9:32 pm
good explanation by palvarez !!
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by Birgit Anne » Mon Nov 30, 2009 11:41 am
I do not understand your last step:

If Model #6 and Model #4 are selected, but Model #5 is not, why is this 4C3?

Is it because there are 4 models left and three of them have to be choosen in order to have a total of 5 models selected?
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by palvarez » Mon Nov 30, 2009 11:49 am
Birgit Anne wrote:I do not understand your last step:

If Model #6 and Model #4 are selected, but Model #5 is not, why is this 4C3?

Is it because there are 4 models left and three of them have to be choosen in order to have a total of 5 models selected?
Yes, 4 left off of 7: 4 and 6 are pre-selected, we dont wanna select 5. so, we are left with remaining 4.

Since we preselected 2, we need to pick remaning 3 from above 4.
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