Permutations and school timetable If there are 6 periods in each working day of a school. In how many different ways can one arrange 5 subjects such that each subject is allowed at least one period? I tried this way- One of the six periods can be arranged in 5 ways and the remaining 5 periods in 5 factorial ways. Totally 600 ways
We assume that, sadly, there are no "spares." So some subject is taught twice, and the others once each.

Choose the subject that gets extra attention. This can be done in $5$ ways. For each such choice, the periods that this favoured subject gets can be chosen in $\binom{6}{2}$ ways, and for each choice the other subjects can be arranged in the remaining slots in $4!$ ways. 

That gives  a total of $5\cdot\binom{6}{2}\cdot 4!$ ways.

Remark: For visualization, I like to think as follos. Imagine that the subjects are A (Art), B (Botany), C, D, and E. A timetable can be thought of as a word of length $6$ over this alphabet, in which one letter (subject)  occurs twice and the remaining letters occur once each. Our task is to count the number of such words. The letter that occurs twice can be chosen in $5$ ways. The $2$ positions it occupies in the word can be chosen in $\binom{6}{2}$ ways, and then the $4$ remaining letters can be put in the vacant positions in $4!$ ways.
There can be 5 cases as the mentioned case below
S1,S2, S3, S4, S5, S1 which can be arranged in 6!/2! ways.
Required no of ways will be 
5× 6!/2! =1800