Possible number of combinations.

1. Aug 8, 2009

dE_logics

I just wanted to confirm this...suppose we need to rearrange n number of objects, the possible ways they can be rearranged is n!

If suppose at the first place, one of the objects cannot be admitted...then it will be (n-1)*(n-1)*(n-2)*(n-3)......right?

Instead of the first place, suppose, an object cannot be admitted to any other place...it wont make any difference right?...I mean, the total combinations will still be (n-1)*(n-1)*(n-2)*(n-3)?

Also related to the above question, if we have n 'types' of objects infinite in quantity (actually it's more of permutation) and one of them cannot be admitted to a place (not necessarily the first)...then the possible number of combinations will be (n-1)*n*n*n...etc...right? Also in case of multiple exceptions for instance object 'x' cannot be admitted to first, second and third place or object 'x' cannot be admitted to first, object 'y' cannot be admitted to second and finally object 'z' cannot be admitted to the third place, then will the possible combinations will be (n-1)*(n-1)*(n-1)*n*n*n.... etc ?

2. Aug 9, 2009

dE_logics

Is it that no one is understanding the question?

3. Aug 9, 2009

daniel_i_l

Yes. The total number of combinations without the constraint is n!. The constraint removes (n-1)! combinations so the total number is:
n! - (n-1)! = n*(n-1)! - (n-1)! = (n-1)(n-1)! = (n-1)(n-1)(n-2)...*2*1
How could it matter?

correct

Last edited: Aug 9, 2009
4. Aug 10, 2009

dE_logics

So I'm right?

Ok, thanks.

Again...thanks.