There are two imposters and seven crewmates on the rocket ‘Plus’. How many ways are there for the nine people to split into three groups of three, such that each group has at least two crewmates? The two imposters and seven crewmates are all distinguishable from each other, but we’re not concerned with the order of the three groups.
For example: \(\{I_1,C_1,C_2\}\), \(\{I_2,C_3,C_4\}\) and \(\{C_5,C_6,C_7\}\) is the same as \(\{C_3,C_4,I_2\}\), \(\{C_5,C_6,C_7\}\) and \(\{I_1,C_2,C_1\}\) but different from \(\{I_2,C_1,C_2\}\), \(\{I_1,C_3,C_4\}\) and \(\{C_5,C_6,C_7\}\).
Draw the plane tiling with regular hexagons.
Can you tile the plane with regular octagons?
Draw how to tile the whole plane with figures, composed from squares \(1\times 1\), \(2\times 2\), \(3\times 3\), \(4\times 4\), and \(5\times 5\) where squares of all sizes are used the same amount of times in the design of the figure.
Show that if \(1+2+\dots+n = \frac{n(n+1)}{2}\), then \(1+2+\dots+(n+1) = \frac{(n+1)((n+1)+1)}{2}\).
Show that \(1+2+\dots+n = \frac{n(n+1)}{2}\) for every natural number \(n\).
Show that if \(1+2^1+2^2+\dots+2^{10} = 2^{11} - 1\), then \(1+2^1+2^2+\dots+2^{11} = 2^{12} - 1\).
Show that \(1+2^1+2^2+\dots+2^n = 2^{n+1} - 1\) for every natural number \(n\).
What is wrong with the following proof that “all rulers have the same length" using induction?
Base case: suppose that we have one ruler, then clearly it clearly has the same length as itself.
Assume that any \(n\) rulers have the same length for the induction hypothesis. If we have \(n+1\) rulers, the first \(n\) ruler have the same length by the induction hypothesis, and the last \(n\) rulers have the same length also by induction hypothesis. The last ruler has the same length as the middle \(n-1\) rulers, so it also has the same length as the first ruler. This means all \(n+1\) rulers have the same length.
By the principle of mathematical induction, all rulers have the same length.