There can be three different cases:
1.A occupies the first position
2.A occupies the last position
3.A occupies a position in between
Here, no two cases can occur at the same time. So, we can find the number of ways in which each case can occur and add them up to find the total number of ways.
Case 1: Suppose A occupies the first place. Then, O cannot occupy the second place.
A × _ _ _ _
O can occupy one of the remaining 4 places. After we fill in O in one of the four positions, we are left with only consonants P, T, R and N. They are allowed to occupy the position next to O. For example, if O occupies the third position, P, T, R and N can occupy the second, fourth, fifth or sixth positions.
A _ O _ _ _
So, the remaining 4 letters can be arranged in the four places in 4! ways. Thus, we can choose a position for O in four ways and, after that, we can arrange the remaining 4 letters in 4! ways. So, the total number of ways in which A can occupy the first position is 4 × 4! = 96.
Case 2: Suppose A occupies the last position. Again, arguing as in case 1, the number of arrangements in this case is 96.
Case 3: Suppose A occupies a position in between. There are 4 possibilities in this case. If A occupies the third position, O cannot occupy the second or fourth positions.
_ × A × _ _
This shows a particular situation where A occupies the third position. So, O can occupy any of the remaining 3 positions. After A and O are allotted positions, there are 4 positions left.
O _ A _ _ _
This shows a particular situation where A occupies the third position and O occupies the first position. The remaining 4 letters can occupy any of the 4 positions in 4! ways. So, the total number of ways in this case 4(A) × 3(O) × 24(P, T, R, N) = 288.
Taking all the three cases together, there are 96 + 96 + 288 = 480 ways in which this can be done.
