A, B, C, D, E and F are cousins. No two cousins are of the same age, but everyone's birthday falls on the same date of the same month. The youngest is 17 years old and the eldest 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?

Let's denote the ages in increasing order: 17, 18, 19, 20, 21, 22.

Given conditions:

1. E is 22 years old. (E = 22)
2. A is one year older than C. (A = C + 1)
3. A is older than B. (A > B)
4. C is older than D. (C > D)
5. F is somewhere between B and D in age. (B D, we have D < C < A. Also, B < A.

Now let's systematically check possible age assignments for C and A, keeping in mind E=22 and all ages must be distinct from 17 to 21 for A, B, C, D, F.

Case 1: (C, A) = (17, 18)

• D < C implies D < 17, which is impossible as 17 is the youngest age.

Case 2: (C, A) = (18, 19)

• D < C implies D = 17 (as D must be less than 18 and distinct).
• B < A implies B < 19. B cannot be 17 (D) or 18 (C). So, no possible age for B. This case is impossible.

Case 3: (C, A) = (19, 20)

• D < C implies D can be 17 or 18.
• B < A implies B < 20.
• If D = 17: C=19, A=20, D=17. B must be < 20 and not 17 or 19. So B = 18. Ages assigned: D=17, B=18, C=19, A=20, E=22. The remaining age for F is 21. Check condition 5 (F is between B and D): Is 21 between 17 and 18? No. This subcase is impossible.
• If D = 18: C=19, A=20, D=18. B must be < 20 and not 18 or 19. So B = 17. Ages assigned: B=17, D=18, C=19, A=20, E=22. The remaining age for F is 21. Check condition 5: Is 21 between 17 and 18? No. This subcase is impossible.

Case 4: (C, A) = (20, 21)

• D < C implies D can be 17, 18, or 19.

• B < A implies B < 21.

• If D = 17: C=20, A=21, D=17. B must be < 21 and not 17 or 20. So B can be 18 or 19.

  • If B = 18: Ages: D=17, B=18, C=20, A=21, E=22. Remaining age for F is 19. Check condition 5: Is 19 between 17 and 18? No. (17 < 19 < 18 is false). This subcase is impossible.
  • If B = 19: Ages: D=17, B=19, C=20, A=21, E=22. Remaining age for F is 18. Check condition 5: Is 18 between 17 and 19? Yes (D < F < B, i.e., 17 < 18 < 19). This gives a valid order: D(17), F(18), B(19), C(20), A(21), E(22). (Order 1)

• If D = 18: C=20, A=21, D=18. B must be < 21 and not 18 or 20. So B can be 17 or 19.

  • If B = 17: Ages: B=17, D=18, C=20, A=21, E=22. Remaining age for F is 19. Check condition 5: Is 19 between 17 and 18? No. This subcase is impossible.
  • If B = 19: Ages: B=19, D=18, C=20, A=21, E=22. Remaining age for F is 17. Check condition 5: Is 17 between 18 and 19? No. This subcase is impossible.

• If D = 19: C=20, A=21, D=19. B must be < 21 and not 19 or 20. So B can be 17 or 18.

  • If B = 17: Ages: B=17, D=19, C=20, A=21, E=22. Remaining age for F is 18. Check condition 5: Is 18 between 17 and 19? Yes (B < F < D, i.e., 17 < 18 < 19). This gives a valid order: B(17), F(18), D(19), C(20), A(21), E(22). (Order 2)
  • If B = 18: Ages: B=18, D=19, C=20, A=21, E=22. Remaining age for F is 17. Check condition 5: Is 17 between 18 and 19? No. This subcase is impossible.

We have found exactly two logically possible orders:

1. D(17), F(18), B(19), C(20), A(21), E(22)
2. B(17), F(18), D(19), C(20), A(21), E(22)

The final answer is B) 2