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probably not

The number of ways by which veena can visit is 4!=24

yes

i can'r go further

*can't

i guess you can list ways in which a comes before B and count them A bit tedious though

yes that is tediuos i want another time saving way

to be honest I hate probability!

sumilarly B before C = 12 ways ?

yes must be

can we multiply these now to get the compound probability?

ok i need to find now the common between A before B and B before C

the question is are they independent events?

i don't understand how independent events means here

(A before B) \(\Large \cap\) (B before C)

by listing and counting there are 4 of them

ok so i need to list them necessrily

i'm afraid i dont remember all the set theory applied to probability

ok so same for 3, 4 and 5 , listing ?

questions iii to v are quite straightforward - i think!!

for q iii it is the number ow ways you can arrange the 3 letters between A and B

= 6 right?

3!=6 , yes

for iv
A first = 3! ways
+ 31 ways with A second

* 3!

ok

v
AB to begin - - 2 ways
AB 2nd and 3rd - 2 waya
AB last - 2 ways

nice question

yea it is

AB--
A_B__
A__B
_AB__
_A_B_
_A__B
__AB
i calculate 7 ways for v.)

|dw:1439460306887:dw| just another way of thinking about it"!

is the 7 ways correct for last one ?

No i dont think so - isn't it A just before B (A and B together?)