If nC4, nC5 and nC6 are in A.P., then n can be:

If a, b, c are in A.P, then 2b = a + c
So applying the same condition, we have
2 × (nC5) = (nC6) + (nC4)
2 × n!/(5!(n-5)!) = n!/(6!(n-6)!) + n!/(4!(n-4)!)
2/(5!(n-5)!) = 1/(6!(n-6)!) + 1/(4!(n-4)!)
2/(5!(n-5)!) = (5!(n-5)!) / (6!(n-6)!) + 1/(4!(n-4)!)
2(6!(n-6)!) = 5!(n-5)! + (5!(n-5)!)(6!(n-6)!)/(4!(n-4)!)
2(6)(5!) = 5!(n-5)! + 6(5!)(n-5)!/((n-4)(n-5)!)
12 = (n-5)! + 6/(n-4)
12(n-4)(n-5)! = (n-4)(n-5)! + 6(n-5)!
12(n-4)(n-5)! = (n-5)!(n-4+6)
12(n-4) = n+2
12n - 48 = n + 2
11n = 50
n = 50/11
This is not an integer solution. Let's recheck our steps.
2 × (nC5) = (nC6) + (nC4)
2 × n!/(5!(n-5)!) = n!/(6!(n-6)!) + n!/(4!(n-4)!)
2/(5!(n-5)!) = 1/(6!(n-6)!) + 1/(4!(n-4)!)
2/(5!(n-5)!) = (n-5)/(6(n-5)!) + (n-3)(n-4)/24
2/120 = (n-5)/720 + (n-3)(n-4)/24
1/60 = (n-5)/720 + (n-3)(n-4)/24
12 = n-5 + 30(n-3)(n-4)
7 = n + 30(n^2 -7n+12)
7 = n + 30n^2 -210n + 360
30n^2 -209n + 353 = 0
Solving the quadratic equation gives approximately n ≈ 11 and n ≈ 10.6 (approximately)
Let's verify for n=14:
14C4 = 1001
14C5 = 2002
14C6 = 3003
2002 = 1001+3003 is false.
Let's check for n=11:
11C4 = 330
11C5 = 462
11C6 = 462
2(462) = 924 ≠ 330 + 462 = 792
Let's try n = 9
9C4 = 126
9C5 = 126
9C6 = 84
2(126) = 252 ≠ 126 + 84 = 210
Let's try n = 12
12C4 = 495
12C5 = 792
12C6 = 924
2(792) = 1584 ≠ 495 + 924 = 1419
Let's try n = 14
14C4 = 1001
14C5 = 2002
14C6 = 3003
2(2002) = 4004 ≠ 1001 + 3003 = 4004
Therefore n = 14