How many solutions are there in positive integers to the equation $x_1 +x_2 +x_3 +x_4 = 16$? Suppose that one of these solutions is chosen at random, with all possibilities being equally likely. What is the probability that the chosen solution has $x_1 = 5$?
For the first part I know it is $15C12 = 455$. For the second part I substituted in $x_1=5$ to get $x_2 +x_3 +x_4 = 11$, which I then transformed into $y_2+y_3+y_4=8$, which has $10C8 = 45$ solutions. This gives a probability of $\frac{45}{455} = \frac{9}{91}$
$(a)$Write an expression for the number of sets $S$ which contain $10$ elements, each of which is an integer between $1$ and $20$. $(b)$ How many subsets of S have size two?$(c)$For a non-empty subset $X ⊆ S$, let $t(X)$ denote the sum of the members of $X$. Prove that there must be distinct subsets $A,B ⊆ S$, each of size two, such that $t(A) = t(B)$.
$(a)$ There are $20C10$ sets. $(b)$ I think this is wrong but I guess $10C2 = 45$. $(c)$ For the 3rd part, the number of subsets of size $2$ must have sums between $3$ and $39$, which is $37$ possible sums. However, there are $45$ possible subsets of size $2$, so by the pigeonhole principle there must exist two subsets with the same sum.