Combinatorics

• November 25th 2013, 10:42 AM
davidciprut
Combinatorics
Hi, I uploaded two statements (I guess), and I didn't understand them , can someone help me understand why do they do k times Q(n-1,k-1) or k times Q(n-1,k)??? I don't understand the k times part..
And the lecturer wrote to the properties of Q, Q(n,1)=1.... shouldn't that be equal to n? Q is just a letter he picked for bijection functions...Does he mean by Q(n,k), n!/(n-k)!??? I hope my questions are clear enogh
• November 25th 2013, 11:15 AM
emakarov
Re: Combinatorics
Quote:

Originally Posted by davidciprut
Q is just a letter he picked for bijection functions...

Do you mean he denotes bijective functions by Q?
• November 25th 2013, 11:42 AM
davidciprut
Re: Combinatorics
Not the bijective functions, the operation of permutation on bijective functions he denoted as Q.. Q(n,k)=n!/(n-k)! I think... because for one to one functions he used A(n,k) but still its the same operation I think...
• November 25th 2013, 12:11 PM
emakarov
Re: Combinatorics
The whole problem seems strange. I am not familiar with the function Q(n, k). A bijection between A and B exists iff n = k. If f is a bijection, then $f^{-1}(f(a))=a$, and $|f^{-1}(f(a))|>1$ is impossible. Not to say that when the vertical bar in the set-builder notation is followed by another bar from an absolute value, the first bar should be changed into, say, a colon.
• November 25th 2013, 12:20 PM
Plato
Re: Combinatorics
Quote:

Originally Posted by davidciprut
Not the bijective functions, the operation of permutation on bijective functions he denoted as Q.. Q(n,k)=n!/(n-k)! I think... because for one to one functions he used A(n,k) but still its the same operation I think...

I fear that you are at the mercy of your instructor, only he/she knows what that means.
I will suggest if you can explain basically what is being asked.
Is it about the number of injections, surjections and/or bijections between sets? Or some variation of that?