Hint for Problem 5: February 2024

First, some hints on the exercises.

List the two-element subsets of \(\{1,2,3,4,5\}\).

The answers are \(1\) and \(n\). It might take a moment of reflection to convince yourself that \(\dbinom{n}{0}=1\) makes sense.

If \(k\) objects are chosen from a set of \(n\) objects, then how many objects are not chosen?

If you are to choose \(k+1\) integers from the set \(\{1,2,3,\dots,n,n+1\}\), then either \(n+1\) is chosen or it is not.

The quantity \((1+x)^n\) is equal to the product of \(n\) copies of \((1+x)\). Try expanding \((1+x)^n\) for a few small values of \(n\) without collecting like terms. As an example, think about how an \(x^{3}\) term could arise during the expansion of \((1+x)(1+x)(1+x)(1+x)(1+x)\).

Below are the hints for the main problems.

If you have never seen a proof of (a)(i), try writing the sum \(1+2+3+\cdots+n\) in reverse order directly under the sum \(1+2+3+\cdots+n\). Now add each column. For (a)(ii), consider the possible values of \(x\). For (a)(iii), consider the equation \(x+y+z=n-1\) for a fixed pair \((x,y)\).

Imagine arranging the \(n\) identical balls in a row and placing \(r-1\) sticks between them. By doing this, you have partitioned the \(n\) balls into \(r\) smaller groups.

Introduce a new variable, \(x_0\), and consider the equation \(x_0+x_1+\cdots+x_r=n\).

The non-negative integers \(x\) with \(x<10^{10}\) are exactly the integers that have at most \(10\) digits. Consider the equation \(x_1+x_2+\cdots+x_{10}=21\) where \(0\leq x_i\leq 9\).

**There was an omission in part (d). The original question said "integers" where it should have said "non-negative integers".**This question can be analyzed by examining the equation \(x_1-x_2+x_3-x_4+x_5-x_6+x_7-x_8=0\). Rearrange this equation and use the ideas from (b) and (d).

Let \(x=a-1\), \(y=b-1\), and \(z=c-1\). Find the number of non-negative integer solutions to \(x+y+z=2024\) with \(x\), \(y\), and \(z\) distinct.