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October 20, 2014

October 20, 2014

Posted by **Jane** on Thursday, October 27, 2011 at 4:43pm.

2) In how many ways can 25 students be assigned to 4 distinguishable study groups if at least 6 students must be in each group?

3) in how many ways can 4 students be assigned to 9 class periods if:

a) there are no restrictions?

b) each students must go to a different period?

- Data management -
**MathMate**, Thursday, October 27, 2011 at 5:18pm1.

There are n=12! ways to line up 12 distinct object into 12 ordered positions.

If positions 1,2,3 form one pile, then N would be overcounted by 3! times. Therefore n must be divided by 3! for each group of 3.

The total number of ways is therefore

12!/(3!3!3!3!)

2. Follow the same argument as in Q1 to place 24 (distinct) students into 4 groups of 6. There are 4 ways to place the 25th student, so multiply by 4.

3.

No restriction:

We give the choice to the students.

Number of choices for the first student=9

number of choices for the second student=9

.....

number of choices for the fourth student=9

Use the multiplication rule to establish the total number of ways.

3.

With restriction that they go to a different session:

Number of choices for the first student=9

Number of choices for the second student = 8

.....

number of choices for the 4th student = 6

Use the multiplication principle to get the total number of arrangements.

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