6 periods and 5 subjects Question : There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period? Is the answer 1800 or 3600 ? I am confused.
Initially this appeared as a simple question. By goggling a bit, I am stuck with two answers. Different sites gives different answers and am unable to decide which is right.
Approach 1 (Source)
we have 5 sub and 6 periods so their arrangement is 6P5 and now we have 1 period which we can fill with any of the 5 subjects so 5C1
6P5*5C1=3600
Approach 2 (Source)
subjects can be arranged in 6 periods in 6P5 ways.
Remaining 1 period can be arranged in 5P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (6P5 x 5P1)/2! = 1800
Alternatively this can be derived using the following approach.
5 subjects can be selected in 5C5 ways.
Remaining 1 subject can be selected in 5C1 ways.
These 6 subjects can be arranged themselves in 6! ways.
Since two subjects are same, we need to divide by 2!
Total number of arrangements = (5C5 × 5C1 × 6!)/2! = 1800
Is any of these approach is right or is the answer different?
 A: Approach 1 is incorrect.  It arranges any five of the courses into one period each, then assigns the left over course to some period.  It double counts each arrangement by having each of the two courses taught at the same time included in the original five.  So the arrangement $12,3,4,5,6$ is counted as $1,3,4,5,6$ plus $2$ in first period and as $2,3,4,5,6$ plus $1$ in first period.  Approach 2 is the same, but acknowledges the double counting in the division by $2!$  
I cannot follow the argument in the paragraph starting "Alternatively" 
A: As Ross Millikan explains, the first approach overcounts and the second approach is a correct version.
An alternative way to do the problem is to notice that there must be exactly one pair of periods with the same subject. You can choose this pair in ${}^6\mathrm C_2=15$ ways. Once you have made this choice, you have five subjects and five places to put them (one of the "places" being the two periods you've paired off). So there are $5!$ ways to do this. Therefore the answer is ${}^6\mathrm C_2\times 5!=1800$
A: i am a student of class 10 and even i had the same question in my book. 
But neither of your answers are correct. The exact answer in 600. I can explain.
_ _ _ _ _ _ imagine these are the 6 space where you should put the 5 subjects.
First 5 periods can be filled in 5P5 ways. i.e. 5!=120 and then the remaining one period can be filled with any one subject from the 5. i.e. 5P1=5.
Then multiply both which is 120*5=600.
And Combination cannot be used because this is a problem based on Permutation.
Combination is mere selection of object. Permutation is an orderly arrangement of object. Here you have to arrange the subjects and not selected.
There you go! 
A: 5 subjects on 5 periods
selecting a place for remaining one subject
Final answer
Over-counting check
