Because we know from the first condition that A cannot finish in the first place, B in the second, etc., we can easily list all the remaining combinations:

BADEC BAECD BCAED BCDEA BCEAD BDAEC BDEAC BDECA BEACD BECAD

BEDAC BEDCA CABED CADEB CAEBD CDAEB CDEAB CDEBA CEABD CEBAD

CEDAB CEDBA DABEC DAEBC DAECB DCAEB DCBEA DCEAB DCEBA DEABC

DEACB DEBAC DEBCA EABCD EADBC EADCB ECABD ECBAD ECDAB ECDBA

EDABC EDACB EDBAC EDBCA

The first condition also eliminates combinations containing one or more of the following pairs of consecutive contestants:

AB BC CD DE

This is what is left:

BEDAC BEDCA ~~CABED~~ ~~CADEB~~ CAEBD ~~CDAEB~~ ~~CDEAB~~ ~~CDEBA~~ ~~CEABD~~ CEBAD

or:

BDAEC BECAD BEDAC BEDCA CAEBD CEBAD CEDBA DAECB DCAEB DCBEA

DCEBA EADCB ECBAD EDACB EDBAC

The second condition requires that exactly two of the contestant finish in the order DAECB.

D A E C B

BDAEC

BECAD

BEDAC

BEDCA *

CAEBD * *

CEBAD

CEDBA

DAECB * * * * *

DCAEB * *

DCBEA *

DCEBA * *

EADCB * *

ECBAD

EDACB * *

EDBAC

This leaves the following six combinations:

CAEBD DCAEB DCEBA EADCB EDACB

We need to find the combination in which two disjoint pairs of students predicted to finish consecutively actually do so. The pairs are: DA AE EC CB

The solution is EDACB, where the two pairs are DA and CB

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