1.If I have a group of 10 seats reserved for people, and there are n=>10 total people, how many ways are there to choose who gets the 10 seats?
for ex:If there was a definite number of people lets say 21 people instead of n>=10 , then it would be 21!/10!11!, but since its n=>10 I cannot seem to figure out the formula for all the results for anything greater or equal to 10.
2.If there a set a(A) of a size i and set b(B) of a size j, i=>j=>1, how would I go about solving how many functions are not one-to-one for f: A->B
for ex:I tried different motions but still getting muffled results(wrong/mixed answers)
3.If there are 30 cards, 6 are red and 7 are black, 10 are not black and red.How many cards are red and black ? --Im assuming it would be along the lines of (see below), am I on the right track or completely off ?
(30) (23) (17)
( 7) (6 ) (10)
^tried styling it didnt work image they are in one bracket and over each other (n/k) type of situation.
4.I have I greater and/or equal to 1, to prove that in any set i+1 ranging from integers (1,2,..,2n) to show that there are going to be 2 consecutive integers.
---I know I need to use pigeonhole principle but I cant budge on the question, I tried implementing it but im positive Im doing it wrong.
Again any guidance would greatly be appreciated.
In general the number is given by $\binom{n}{10}=\frac{n!}{10!(n-10)!}$.
I suggest you count the total number of functions, and subtract the number of one-to-one functions. (There are formulas for both of these numbers.)
If, of the $30$ cards, we have that "10 are not black and red" then $30-10$ are "red and black", right? (Is this phrased exactly how the question was asked?)
We have a set $S$ of $n+1$ elements from $\{1,2,\ldots,2n\}$. We set up $n$ pigeonholes $\{1,2\}$, $\{3,4\}$ and so on up to $\{2n-1,2n\}$. At least one of the pigeonholes has both elements in $S$.