1.
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.
1) In how many ways can 12 distinct objects be separated into three equal piles?
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?
1 answer
1 answer