If there are six period in each working day of a school , in how many ways can one arrange 5 subject such that each subject is allowed at least one period ?
To solve this problem, we can use the concept of stars and bars.
Let's say there are 5 subjects: A, B, C, D, and E. We want to arrange these subjects in 6 periods such that each subject has at least one period.
To ensure that each subject has at least one period, we can first distribute one period to each subject. So each subject now has 1 period assigned to it, and we are left with 1 additional period to distribute.
Now, we can use stars and bars to distribute the remaining period among the subjects. We can think of the additional period as a "star," and the areas between the subjects as "bars" that separate them. We need to distribute the star among the bars in order to assign the remaining period.
We have 5 subjects and 6 periods, so there are 5-1=4 bars. We need to distribute the star (additional period) among these 4 bars.
Using the stars and bars formula, the number of ways to distribute the star among the bars is given by (n+k-1) choose (k-1), where n is the number of stars and k is the number of bars.
In this case, we have 1 star and 4 bars, so the number of ways to distribute the periods is (1+4-1) choose (4-1) = 4 choose 3 = 4.
Therefore, there are 4 ways to arrange the 5 subjects such that each subject is allowed at least one period.
To solve this problem, we can use the concept of permutations with restrictions.
First, let's calculate the total number of ways to arrange the 5 subjects in 6 periods without any restrictions. We can do this by finding the number of permutations:
6P5 = 6! / (6-5)! = 6! / 1! = 6 * 5 * 4 * 3 * 2 * 1 = 720
However, this includes arrangements where one or more subjects are not assigned any periods. We need to subtract these cases to satisfy the restriction that each subject should have at least one period.
There are 5 subjects, so we can choose one subject to be left out in 5 ways. Once we have chosen a subject, we arrange the remaining 4 subjects in the 6 periods in a way that satisfies the restriction. This can be done in (6P4) = 6! / (6-4)! = 6! / 2! = 6 * 5 * 4 * 3 = 360 ways.
Therefore, the number of ways to arrange 5 subjects with each subject having at least one period is given by:
Total number of ways - Number of ways with one subject left out = 720 - 360 = 360 ways.
Hence, there are 360 ways to arrange the 5 subjects under the given restrictions.
To find the number of ways to arrange 5 subjects such that each subject is allowed at least one period in a day with 6 periods, we can use a combination of counting techniques.
Step 1: Choose one subject for each period:
Since each subject is allowed at least one period, we can assign one subject to each of the six periods in 5! (5 factorial) ways.
Step 2: Assign the remaining 5 subjects to the periods:
Now, we have 5 subjects left to assign to the 6 periods. We can use the stars-and-bars method to find the number of ways to distribute these subjects.
Consider the analogy of 5 subjects (stars) and 6 periods (bars). The 5 subjects need to be placed independently between the bars, representing each period.
Using the formula for stars and bars, we have (n + r - 1) choose (r - 1), where n is the number of objects (subjects) and r is the number of bins (periods).
In this case, n = 5 (number of subjects) and r = 6 (number of periods). Plugging these values into the formula, we get (5 + 6 - 1) choose (6 - 1) = 10 choose 5 = 252.
Step 3: Multiply the results from Step 1 and Step 2:
To find the total number of arrangements, we multiply the number of ways to assign one subject to each period (5!) by the number of ways to assign the remaining subjects (252).
Total number of arrangements = 5! * 252 = 120 * 252 = 30,240.
Therefore, there are 30,240 ways to arrange 5 subjects such that each subject is allowed at least one period in a day with 6 periods.