Question

A certain classroom has two rows of seats. The front row contains 8 seats and the back row contains 10 seats. How many ways are there to seat 15 students if a certain group of 4 of them refuses to sit in the back row and if a certain group of 5 others refuses to sit in the front row?

My approach:
4 has to go front and 5 has to go back.
So I splited them into 4 groups

1) 4 front 4 others / 5 back 2 others
2) 4 front 3 others / 5 back 3 others
3) 4 front 2 others / 5 back 4 others
4) 4 front 1 others / 5 back 5 others

However, I cannot put them into equations.

Additionally, if anyone knows website that has many combination problems with detailed solutions, please let me know. Websites that I found has only very basic information.

Thanks in advance.

Answer 1

You can consider the three groups of students separately.

• For the group that has to sit in the front row, there are 8 Perm 4
  different possible places for them to sit.
• For the group that has to sit in the back row, there are 10 Perm 5 different
  possible places for them to sit.
• For the remaining 6 students, there will always be 18 - 4 - 5 = 9 seats left
  for them to choose from, hence a total of 9 Perm 6 choices.

All together this yields (8!/4!)(10!/5!)(9!/3!) = 3072577536000.

Note: This is eerily similar to Problem 14 from Chapter 3 in R. Brualdi, Introductory combinatorics, is this for a homework?