Puzzles

The final digit is 1

Once the leftmost digit of a row and the row below it are decided, there is only one way to fill the rest of the row. Hence, there are two options for the next row (0 0 or 1 1), and four options for the next row (the rows starting 0 and 1 for both 0 0 and 1 1), and in general there are double the number of options for each row as we go upwards.

Therefore for the row with 10 digits there are 2⁹ = 512 options.

To make sequences of 4 numbers, we can insert 4s into three different places in the length 3 sequences given to obtain:

There are some possibilities missing: those containing \(i, 4, i+1\). These can be found by taking the sequences of length 2, picking a number \(i\), adding 1 to every number larger than \(i\), then replacing \(i\) with \(i\ 4\ i+1\).

This gives a total of 3×3+2×1=11 sequences for 4 numbers.

To make sequences with 5 numbers, we can insert 5s into four different places in the length 4 sequences. This gives 4×11=44 sequences.

The missing sequences can then be found by taking the sequences of length 3, then doing the same process as above:

There are 3×3=9 of these, giving 44+9 = 53 total sequences of length 5.

To make sequences with 6 numbers, we can insert 6s into five different places in the length 5 sequences. This gives 5×53=265 sequences. We can also make sequence by picking a number to replace in the length 4 sequences. This gives 4×11=44 more sequences. Therefore there are 265+44 = 309 sequences in total.

There are 9 gaps between the numbers 1 to 10. We need to pick 5 of these gaps to split the sequence. There are \(\left(\begin{array}{c}9\\5\end{array}\right)\) ways to do this: this is 126.

154

328