1. ## permuations

On Sunday you can either go to the cinema or to the the theatre, on
each of Monday, Tuesday and Wednesday you can study either maths,
physics or chemisctry, on each of Thursday, Friday and Saturday you
can visit one of your 5 best friends (you must choose exactly one thing
for each day).
In how many different ways can you plan your week?

proposed solution:
1 + 3! + 5!/(5-3)!

2. I think you need a product of seven factors, one for each day.

3. Originally Posted by baz
On Sunday you can either go to the cinema or to the the theatre, on
each of Monday, Tuesday and Wednesday you can study either maths,
physics or chemisctry, on each of Thursday, Friday and Saturday you
can visit one of your 5 best friends (you must choose exactly one thing
for each day).
In how many different ways can you plan your week?

proposed solution:
1 + 3! + 5!/(5-3)!
You titled this "permutations," but since you can have duplicate activities on subsequent days this is not really a permutations problem. Like emakarov said, this problem should reduce to a large product, and I think, no factorials.

Edit: It strikes me that "you must choose exactly one thing for each day" is ambiguous. I assumed it meant, you can't study both maths and physics on Monday. But perhaps it's intended to convey, you can't study physics on both Monday and Tuesday. In the latter case we would be thinking of binomial coefficients, and if the number of choices matches the number of days, this reduces to a simple factorial.

4. Originally Posted by emakarov
I think you need a product of seven factors, one for each day.
Emakarov,what seven factors are you talking about?

5. Originally Posted by undefined
You titled this "permutations," but since you can have duplicate activities on subsequent days this is not really a permutations problem. Like emakarov said, this problem should reduce to a large product, and I think, no factorials.

Edit: It strikes me that "you must choose exactly one thing for each day" is ambiguous. I assumed it meant, you can't study both maths and physics on Monday. But perhaps it's intended to convey, you can't study physics on both Monday and Tuesday. In the latter case we would be thinking of binomial coefficients, and if the number of choices matches the number of days, this reduces to a simple factorial.
yes,I also think it is not permutation problem.Can it be solved through inclusion-exclusion principle?

6. Originally Posted by baz
yes,I also think it is not permutation problem.Can it be solved through inclusion-exclusion principle?
There is no possibility of overlap, so inclusion-exclusion is not indicated.

Under my first interpretation of the problem, we simply multiply all possibilities.

(2^1)*(3^3)*(5^3) = whatever

Under the second interpretation, C(n, k) means n choose k,

(2)*(3!)*(C(5, 3)) = whatever

If you want to be thorough, just provide both answers and say the question is ambiguous.

(Edited out wrong formula for first interpretation.)

EDIT: I did the second interpretation wrong now, haha. You can choose 3 friends out of 5 in C(5,3) ways; you can permute them throughout the days in 3! ways. So the expression would be

(2)*(3!)*(C(5, 3)*3!) = result

Sorry about that. The explanation I gave in my previous post was off. If you wanted to write it out very explicitly, you write

(C(2,1)*1!)*(C(3,3)*3!)*(C(5,3)*3!)

Hope it's clear.

7. EDIT: I did the second interpretation wrong now, haha. You can choose 3 friends out of 5 in C(5,3) ways; you can permute them throughout the days in 3! ways. So the expression would be

(2)*(3!)*(C(5, 3)*3!) = result

[/quote]
When we are choosing 3 friends out of three it means we are permuting them on three days,I think no need to multiply with 3!.
We can't use product rule,we would have use tat had there been no restriction of doing activities on particular days of week.I hope you get it.
I think sum is appropriate.
This is ricky question.

8. Originally Posted by baz
When we are choosing 3 friends out of three it means we are permuting them on three days,I think no need to multiply with 3!.
We can't use product rule,we would have use tat had there been no restriction of doing activities on particular days of week.I hope you get it.
I think sum is appropriate.
This is ricky question.

As for the permutations, I believe my answer stands. Consider the smaller example:

You have three friends. On Monday and Tuesday, you visit two different friends, one on each day. How many ways can you do this?

{1,2,3} represents the friends

You have

(1,2) (2,1)
(1,3) (3,1)
(2,3) (3,2)

which represents all possible ways.

C(3,2) = 3

2! = 2

The number of combinations is indeed C(3,2)*2! = 6.

9. Originally Posted by undefined

As for the permutations, I believe my answer stands. Consider the smaller example:

You have three friends. On Monday and Tuesday, you visit two different friends, one on each day. How many ways can you do this?

{1,2,3} represents the friends

You have

(1,2) (2,1)
(1,3) (3,1)
(2,3) (3,2)

which represents all possible ways.

C(3,2) = 3

2! = 2

The number of combinations is indeed C(3,2)*2! = 6.
Ok I got it.3! needs to be multiplied with c(5,3).

By sum I mean 2! + 3! + c(5,3).3!

10. Originally Posted by baz
Ok I got it.3! needs to be multiplied with c(5,3).

By sum I mean 2! + 3! + c(5,3).3!
I don't see how that matches up at all. Try doing sanity checks, meaning try to detect when you get an answer that doesn't make sense. (A general example of a failed sanity check is if you find a distance that is negative, or a non-integer number of people.)

You can list out all combinations when small numbers are involved, and see if your formula gives the right answer. If it passes the sanity check, it might be right. If it fails the sanity check, then you can rule it out as wrong.