**Explanation:**

Ammonia (NH3) is a molecule composed of one nitrogen atom and three hydrogen atoms. In order to determine the number of IR active vibrational modes in ammonia, we need to consider its molecular symmetry and the number of atoms present.

**Molecular Symmetry:**

Ammonia belongs to the point group C3v, which has a threefold rotation axis (C3) passing through the nitrogen atom and a vertical mirror plane (σv) containing the nitrogen and hydrogen atoms.

**Vibrational Modes:**

Vibrational modes can be classified based on their symmetry properties. In general, there are three types of vibrational modes:

1. Stretching modes: These modes involve the stretching of bonds and can be further classified into symmetric stretching (ν1), asymmetric stretching (ν2), and degenerate stretching (ν3).

2. Bending modes: These modes involve the bending of bonds and can be further classified into in-plane bending (ν4) and out-of-plane bending (ν5).

3. Torsional modes: These modes involve the twisting or rotation around a bond.

**Application of Group Theory:**

Group theory can be used to determine the number of IR active vibrational modes in a molecule. In the case of ammonia, the C3v point group has three irreducible representations: A1, A2, and E.

**Application of Selection Rules:**

Selection rules help us determine which vibrational modes are IR active. In general, for IR active modes, the change in dipole moment (∆µ) should be non-zero during the vibration.

For ammonia, the selection rules are as follows:

- A1 modes: ∆µ = 0
- A2 modes: ∆µ = 0
- E modes: ∆µ ≠ 0

**Determination of IR Active Modes:**

By applying the selection rules, we find that the A1 and A2 modes are not IR active since they have ∆µ = 0. However, the E modes have ∆µ ≠ 0, indicating that they are IR active.

In ammonia, there are three E modes: ν2 (in-plane bending), ν3 (degenerate stretching), and ν4 (out-of-plane bending). Therefore, the number of IR active vibrational modes in ammonia is 3.

Hence, the correct answer is option A (6 is not the correct answer).