Here, sample space will contain 4! ways.
S=
{ABCD,ACBD,ABDC,ADBC,ACDB,ADCB,
BACD,BADC,BCAD,BCDA,BDAC,BDCA,
CABD,CADB,CBAD,CBDA,CDAB,CDBA,
DABC,DACB,DBAC,DBCA,DCAB,DCBA}
So, n(s)=24
(i)Number of ways Veena travels A before B =12
∴ Probability of her traveling A before B= 12/24 = 1/2
(ii) Number of ways Veena travels A before B and B before C= 4
:. Probability of her traveling A before B and B before C= 4/24 = 1/6
(iii)Number of ways Veena travels A first and B last= 2
:. Probability of her traveling A first and B last = 2/24 = 1/12
(iv)Number of ways Veena travels A either first or second = 12
:. Probability of her traveling A either first or second= 12/24 = 1/2
(v)Number of ways Veena travels A just before B = 6
:. Probability of her traveling A just before B = 6/24 = 1/4
