Hmmm, well, maybe this is the 2nd tough combination problem from PR.

Let's arrange people from shortest to tallest:

A-B-C-

D-E-

F-G

4th and 6th tallest persons are D and F in red

I proceeded with favorable outcome approach

1- If D and F are in the group of 5, then E has to be there so that D and F be separate as well as the group be arranged from shortest to tallest. Hence, if we have D, E and F in the group of 5, we have 4C2 ways of selecting the 2 remaining people{A,B,C,G} to form the group of 5. 4C2 =

6
2- If only D is in the group of 5 and F is to be excluded, we have 5C4 ways of selecting 4 people from the group made up of A,B,C,E or G. Remember that F is already removed and should not be among the sample group. 5C4 =

5
3- If only F is in the group of 5 and D is to be excluded, it is the same process as in step 2 above. 5C4 =

5
Answer is sum of all those: 6+5+5 = 16

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Best Regards,

Paul