# Help me, to solve this ?? [closed]

Question:

Show that:

1. P(n,n) = n!

2. C(n,0) = 1

3. C(n,n) = 1

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What have you tried? How have you had those functions defined? –  Tobias Kildetoft Apr 1 at 14:16
For those like me: $P(n,k)$ is defined here. –  julien Apr 1 at 14:26
@julien Thanks, I had no idea what it was either. –  Git Gud Apr 1 at 14:28
When I saw $C(n,n)$ I thought Stirling numbers of the second kind. –  MITjanitor Apr 1 at 14:56

## closed as not a real question by Andres Caicedo, Stefan Hansen, Dennis Gulko, Arkamis, MicahApr 1 at 16:15

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How many subsets of size $0$ are there in $\{1,\dots,n\}$? How many of size $n$?

In how many ways can you change the order of the elements in $\{1,\dots,n\}$ so that it "stays the same"? For example $$\{1,2,3\}=\{3,1,2\}=\{1,3,2\}$$

Hint You have to choose the first element out of $n$. Once you did this, you are left with $n-1$ elements to choose from. Continuing, you have to exhaust all $n$ elements, so you get $n(n-1)(n-2)\dots=\text{ ? }$

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Without using formulas, assuming that you are given the usual definitions for P and C.

P(n,n) is the number of permutations of a set of n distinct objects. Let us count these permutations. First, there are n choices for the first object, (n-1) for the second, and so forth. This yields n!

C(n,0) is the number of ways to select 0 objects from a set of n objects. Thus you must not select the first object, you must not select the second object, and so forth, until you must not select the nth object. There is only one choice for each selection, so multiply them all up, and there is only 1 way to do it.

C(n,n) is the same, just remove the word "not" from the above paragraph.

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you can use from the formula too . $p(n,n)=\frac{n!}{(n-n)!}=n!$ $C(n,o)=\frac{n!}{(n-0)! 0!}=1$ $C(n,n)=\frac{n!}{(n-n)! n!}=1$