What is the probability that the dj will play 3 songs in any order that is not consecutive? Every Monday morning from 0700-0800 hours, the radio disk jockey (dj) on duty
must play 8 songs on radio; the songs are pre-selected by his manager and the dj may
play any eight of them randomly. 
The composition of songs selected for next Monday follows: 5 Jazz, 2 Soul, 10 R & B and 5 Country songs.
What is the probability that, on Monday next week, the dj will play three R & B songs in any order that is not consecutive?
I understand that we are choosing 3 songs that are R&B and then the rest but I'm getting confused with how to sort the problem without involving consecutive order.
The answer they gave : 0.1061
 A: There are Perm(22,8) possible song orders if there are no restrictions.  To count the number of desirable song orders:
(a) Choose 3 R&B songs. (Order unimportant)
(b) Choose 5 other songs (Order unimportant)
(c) Choose 3 slots of 8 with no two consecutive (for the R&B songs).  (This can be counted by 'brute force', or by some sort of 'stars and bars' method)
(d) Choose an order for the R&B songs into their designated slots
(e) Choose an order for the other 5 songs into their designated slots.
A: My attempt at this:
No. of ways of choosing 3 R&B and 5 non-R&B songs, let us say is $A = \binom{10}{3} \times \binom{12}{5}$
For each of the $A$ combinations, there are $8!$ unique sequences. 
Out of these sequences, exclude those that satisfy the consecutive R&B condition. For this, visualise $3$ R&B songs joined together back to back as one song ($3!$ ways) and played together with the $5$ other songs ($6!$ ways). So total no. of sequences satisfying consecutive condition is $3! \times 6!$ ways. 
So total no. of sequences acceptable (i.e., not satisfying consecutive condition) is $  \binom{10}{3} \times \binom{12}{5} \times (8! - 3! \times 6!)$
Total no. of all sequences is $_{22}P_8$
Probability is then 
$  \frac{\binom{10}{3} \times \binom{12}{5} \times (8! - 3! \times 6!)}{_{22}P_8}$
