Combination & Permutation Problem

Hello I need help with part c & d of this question... I only have answers for these.
The answer to c is 39,916,800 and for d the answer is 59,512,320.

My workings for c:
I know that there are two situations.
I can select 6 blue chairs & 3 green chairs.
OR I can select 5 blue chairs and 4 green chairs.

The number of ways that I can select 6 blue chairs & 3 green chairs would be 9P6x6P3=7,257,600 --> the number of ways of selecting 6 people from 9 people and arranging them in the 6 blue chairs and multiplying that by the number of ways of selecting 3 chairs from 6 green chairs and arranging three people in those 3 green chairs.

I think the number of ways that I can select 5 blue chairs & 4 green chairs would be 6P5x6P4=259,200 --> number of ways of selecting 5 blue chairs from 6 blue chairs and arranging 5 people in those blue chairs and multiplying that by the number of ways of selecting 4 chairs from 6 green chairs and arranging 4 people in those 4 green chairs!

And when I add 7,257,600 and 259,200 my answer is wrong..haha
What am I doing wrong?

My workings for part d:

4x11P9 + 4C2x10P9 + 4C3x9P9

number of ways of having one chair unoccupied + two chairs unoccupied + 3 chairs unoccupied

and yes it gives a wrong answer.

Please help me.

To begin with the 5,4 allocation for c), try and be a bit clearer about what is selected/not selected

I simply wrote out the numbers rather than try and use nPr/nCr formula so for blue gives
9*8*..*5
ways of seating 5 people in 5 blue seats, then how many ways of arranging the free blue seat.

Similar for green. Then combine.

I generally find drawing stuff/doing a simplified problem helps. There are a few ways to approach it (arguably neater/simpler mathematically), but getting an intuitive solution working first is important.

So for part c I did 6P5 x 6P4=259,200, and by this I meant 6C5 x 5! x 6C4 x 4!. I selected 5 blue chairs from 6 blue chairs and arranged 5 people who would sit on the 5 blue chairs- this is 6C5 x 5!. Then I selected 4 green chairs from 6 green chairs and arranged 4 people who would sit on the 4 green chairs- this is 6C4 x 4!. And I multiplied the two values.

You wrote: 'I simply wrote out the numbers rather than try and use nPr/nCr formula so for blue gives 9*8*..*5 ways of seating 5 people in 5 blue seats, then how many ways of arranging the free blue seat.'

Thank you for explaining it, but I don't quite understand what you did here ... Why is the number of ways of seating 5 people in 5 blue seats 9*8*..*5?
This thing is so confusing for me.

There are a few ways to view it, but each chair is unique so order matters, so if there are 5 (blue) chairs to fill, then 9 choices for the first, 8 for the second ... 5 for the fifth. So given that there is a spare (blue) chair, there are
9*8*...*5*6
ways of allocating 5 (out of 9) people to 5 (out of 6) blue chairs. Obviously, this can be given an nCr/nPr representation.

So similar for the other 4 people to 4 out of 6 green chairs, then combine.

If youve done part c) correctly you should see its basically
9! * #arrangements of 3 empty squares
The number of different arrangements of 3 empty squares should help you then get d).

For the 5,4 allocation for c), I tried 9P5 x 6C5 + 6C4 x 4! based on what you told me but the number is too small lol. What am I doing wrong?
What I did:
1. Choose 5 blue chairs from the 6 blue chairs (6C5).
2. Select 5 people from 9 people and arrange them in the 5 selected blue chairs (6C5).
3. Select 4 green chairs from the 6 green chairs (6C4)
4. Arrange the remaining 4 people in the 4 green chairs (4!)

You seem to still be thinking too much about nCr/nPr formula rather than sketching / working through a simpler problem. With these question types, this will often cause problems.

Your first, third and fourth points are correct. Note 6C5 = 6C1 and 6C4 = 6C2.
* The 9 people can be put in the first blue seat, 8 in the second blue, ... 5 in the fifth blue. Your second point (bold) is not correct.

Why are you adding (bold)?

Oh wait- the second point should be 9P5, which is 9*8*7*6*5. I wrote the wrong thing.

Agreed . But why add the different colours?

Ohh I multiplied them and it works! I'm finally done with part c).

I tried part d) by calculating 4C1*11P9 + 4C2*10P9 + 4C3*9P9 but it doesn't work.
What I did:
1. Considered the case where 1 chair in row 2 is unoccupied (4C1). Since one chair will be unoccupied, I exclude that chair and select 9 chairs from 11 chairs (11C9). Then I arrange 9 people in the selected 9 chairs (9!).
2. Considered the case where 2 chairs in row 2 is unoccupied (4C2). Since 2 chairs will be unoccupied, I exclude those chairs and select 9 chairs from 10 chairs (10C9). Then I arrange 9 people in the selected 9 chairs (9!).
3. Considered the case where 3 chairs in row 2 is unoccupied (4C3). Since 3 chairs will be unoccupied, I exclude those chairs and select 9 chairs from 9 chairs (9C9). Then I arrange 9 people in the selected 9 chairs (9!).

I added 1,2,3 altogether but the answer is wrong. It should be 59,512,320. What am I doing wrong?