The answer is 48. Say I represent the seats like this:

{a} <--- Row 3

{b} {c}{d}{e} <--- Row 1

Betty and Veronica cannot sit in seat {b} or sit next to each other: so there are 4 scenarios to consider that fulfill these conditions:

1) Betty in seat {a} and Veronica in seat's {c}, {d}, or {e}
2) Veronica in seat {a} and Betty in seat's {c}, {d}, or {e}
3) Veronica in seat {c} and Betty in seat {e}
4) Betty in seat {c} and Veronica in seat {e}

So to find the total number of seating arrangements we have to find the number of arrangements in cases 1-4 and then add them up!

1) One choice for seat {a} (Betty), three choices for seat {b} (Archie, Jerry, or Moose) and 3! choices for the remaining seats {c} {d} and {e}.

So the total number of arrangements in case 1) is: 1x3x3! = 18

2) The only difference here is that we've placed Veronica in seat {a} instead of Betty,

So the total number of arrangements is the same as in 1) = 1x3x3! = 18

3) If Veronica is in seat {c} and Betty is in seat {e} then as usual we have 3 choices for seat {b}, two choices for seat {d} and one choice for seat {a} because we're seating each of the three boys in the remaining seats.

So the total number of arrangements is 3x2x1 = 6

4) This scenario is exactly the same as scenario 3) except that we've switched Betty and Veronica's seat, this time Betty is in seat {c} and Veronica is in seat {e}.

So the number of arrangements is the same as in case 3) which is 3x2x1 = 6

Adding up the number of arrangements in each case gives us the total number of possible arrangements:

18 + 18 + 6 + 6 = 48

I hope this helps!

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