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.
probability problem with condition
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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.
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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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?
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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Yes, 4 left off of 7: 4 and 6 are pre-selected, we dont wanna select 5. so, we are left with remaining 4.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?
Since we preselected 2, we need to pick remaning 3 from above 4.