There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular-skilled employees (RE). During the next five months, the division has to complete five projects every month. Out of the 25 projects, 5 projects are "challenging", while the remaining ones are "standard". Each of the challenging projects has to be completed in different months. Every month, five teams - T1, T2, T3, T4 and T5, work on one project each. T1, T2, T3, T4 and T5 are allotted the challenging project in the first, second, third, fourth and fifth month, respectively. The team assigned the challenging project has one more employee than the rest.

In the first month, T1 has one more SE than T2, T2 has one more SE than T3, T3 has one more SE than T4, and T4 has one more SE than T5. Between two successive months, the composition of the teams changes as follows:

a. The team allotted the challenging project, gets two SE from the team which was allotted the challenging project in the previous month. In exchange, one RE is shifted from the former team to the latter team.

b. After the above exchange, if T1 has any SE and T5 has any RE, then one SE is shifted from T1 to T5, and one RE is shifted from T5 to T1. Also, if T2 has any SE and T4 has any RE, then one SE is shifted from T2 to T4, and one RE is shifted from T4 to T2.

Each standard project has a total of 100 credit points, while each challenging project has 200 credit points. The credit points are equally shared between the employees included in that team.

The number of SE in T1 and T5 for the projects in the third month are, respectively:

The correct option is A.

First, let us find the size of the teams. Let the standard team size be x. Since the team with the challenging project has one more employee than the rest, its size is x + 1. Since there are 21 employees in total:
4x + (x + 1) = 21
5x = 20
x = 4
Thus, standard teams have 4 employees, and the challenging project team has 5 employees.

In Month 1, T1 is the challenging team (size 5), and T2, T3, T4, T5 are standard teams (size 4). Let the number of SEs in T1, T2, T3, T4, T5 in Month 1 be S₁, S₂, S₃, S₄, S₅ respectively. We are given:
S₁ = S₂ + 1 = S₃ + 2 = S₄ + 3 = S₅ + 4
Since the total number of SEs is 10:
(S₅ + 4) + (S₅ + 3) + (S₅ + 2) + (S₅ + 1) + S₅ = 10
5S₅ + 10 = 10 ⇒ S₅ = 0.
Thus, the distribution of (SE, RE) in Month 1 is:
T1: (4, 1)
T2: (3, 1)
T3: (2, 2)
T4: (1, 3)
T5: (0, 4)

Now let's trace the transitions to Month 2:
Rule a: T2 (challenging team in Month 2) gets 2 SE from T1 (challenging team in Month 1). In exchange, 1 RE shifts from T2 to T1. Before Rule b, the compositions are:
T1: (2, 2)
T2: (5, 0)
T3: (2, 2)
T4: (1, 3)
T5: (0, 4)
Rule b:
1) T1 has 2 SE (> 0) and T5 has 4 RE (> 0). Thus, 1 SE shifts T1 → T5 and 1 RE shifts T5 → T1. T1 becomes (1, 3) and T5 becomes (1, 3).
2) T2 has 5 SE (> 0) and T4 has 3 RE (> 0). Thus, 1 SE shifts T2 → T4 and 1 RE shifts T4 → T2. T2 becomes (4, 1) and T4 becomes (2, 2).
So the final composition in Month 2 is:
T1: (1, 3), T2: (4, 1), T3: (2, 2), T4: (2, 2), T5: (1, 3)

Now let's trace the transitions to Month 3:
Rule a: T3 (challenging team in Month 3) gets 2 SE from T2. In exchange, 1 RE shifts from T3 to T2. Before Rule b, the compositions are:
T1: (1, 3)
T2: (2, 2)
T3: (4, 1)
T4: (2, 2)
T5: (1, 3)
Rule b:
1) T1 has 1 SE (> 0) and T5 has 3 RE (> 0). Thus, 1 SE shifts T1 → T5 and 1 RE shifts T5 → T1. T1 becomes (0, 4) and T5 becomes (2, 2).
2) T2 has 2 SE (> 0) and T4 has 2 RE (> 0). Thus, 1 SE shifts T2 → T4 and 1 RE shifts T4 → T2. T2 becomes (1, 3) and T4 becomes (3, 1).
So the final composition in Month 3 is:
T1: (0, 4)
T2: (1, 3)
T3: (4, 1)
T4: (3, 1)
T5: (2, 2)

Thus, in the third month, the number of SE in T1 and T5 are 0 and 2, respectively.