A, B, C, D, E and F are cousins. No two cousins are of the same age, but all have birthdays on the same day of the same month. The youngest is 17 years old and the oldest E is 22 years old. F is somewhere between B and D in age. A is older than B. C is older than D. A is one year older than C.

What is the number of logically possible orders of all six cousins in terms of increasing age?

The correct option is 2.

Let's analyze the ages of the six cousins: A, B, C, D, E, and F. We are given the following conditions:

1. No two cousins are of the same age, and they all have birthdays on the same day. Since the youngest is 17 years old and the oldest, E, is 22 years old, the ages of the six cousins must be exactly the set of integers from 17 to 22:
{17, 18, 19, 20, 21, 22}.
Thus, E is 22 years old.

2. F is somewhere between B and D in age. This means either:
$B<F<D$
or
$D<F<B$.

3. A is older than B ($A>B$).

4. C is older than D ($C>D$).

5. A is one year older than C ($A=C+1$).

From condition 5, A and C must occupy consecutive ages (with A being older). Since E is 22, the possible ages for the pair (C, A) can be:
- Case 1: C is 20 and A is 21.
- Case 2: C is 19 and A is 20.
- Case 3: C is 18 and A is 19.
- Case 4: C is 17 and A is 18.

Let's evaluate each case to see which ones are logically consistent with the other constraints:

Case 1: C = 20, A = 21
Since E = 22, the remaining ages for B, D, and F are {17, 18, 19}.
- We are given $C>D$, so D must be younger than 20. Since the remaining ages are all less than 20, this condition is satisfied for any choice.
- We are given $A>B$, so B must be younger than 21, which is also satisfied.
- F is between B and D. Since B, D, and F must occupy the set {17, 18, 19}, F must be the middle value, which is 18. This gives us two possibilities for B and D:
- Subcase 1a: B = 17, F = 18, D = 19. Here, $C>D$ (20 > 19) is satisfied. The full order from youngest to oldest (17 to 22) is: B, F, D, C, A, E.
- Subcase 1b: D = 17, F = 18, B = 19. Here, $C>D$ (20 > 17) is satisfied. The full order from youngest to oldest is: D, F, B, C, A, E.

Case 2: C = 19, A = 20
Since E = 22, the remaining ages for B, D, and F must come from {17, 18, 21}.
- Since $A>B$, we must have $20>B$. Thus, B cannot be 21. B must be 17 or 18.
- Since $C>D$, we must have $19>D$. Thus, D cannot be 21. D must be 17 or 18.
- This leaves 21 as the only remaining age, which must be assigned to F. However, F must be between B and D. Since B and D are both in {17, 18}, F (21) cannot be between them. Thus, Case 2 is impossible.

Case 3: C = 18, A = 19
The remaining ages for B, D, and F must come from {17, 20, 21}.
- Since $C>D$, we must have $18>D$, so D must be 17.
- This leaves B and F to be 20 and 21 in some order.
- Since D = 17, F (either 20 or 21) cannot be between B and D because there is no remaining age smaller than 17 to place B on the other side of F. Thus, Case 3 is impossible.

Case 4: C = 17, A = 18
- Since $C>D$ and C is the youngest (17), there is no age left for D that is younger than C. Thus, Case 4 is impossible.

Consequently, there are only 2 logically possible orders of all six cousins in terms of increasing age (B, F, D, C, A, E and D, F, B, C, A, E).