It is given 3 numbers : 1, 3, and 5, you were told to write numbers by adding those 3 numbers, for example: There are 8 ways of writing the number 6
6 = 1 + 5
6 = 5 + 1
6 = 3 + 3
6 = 1 + 1 + 1 + 3
6 = 1 + 1 + 3 + 1
6 = 1 + 3 + 1 + 1
6 = 3 + 1 + 1 + 1
6 = 1 + 1 + 1 + 1 + 1 + 1
How many ways are there to write the number 12. The problem is, I need to finish it quick. I can do it manually, but it'll take some time. Can anybody help me? Just giving hints would help a lot! Thanks
There are three distinct ways to write 12 as the sum of 1s, 3s and 5s:
1) Write 11 as the sum of 1s 3s and 5s and then add an extra 1
2) Write 9 as the sum of 1s, 3s and 5s and then add an extra 3
3) Write 7 as the sum of 1s, 3s and 5s and then add an extra 5
So if $f(n)$ is the number of ways of writing $n$ as the sum of 1s, 3s and 5s then we have
$f(12) = f(11) + f(9) + f(7)$
But by a similar argument we know that $f(11) = f(10) + f(8) + f(6)$. So
$f(12) = f(11) + f(9) + f(7) \\=f(10)+f(9)+f(8)+f(7)+f(6) \\=2f(9)+f(8)+2f(7)+f(6)+f(5) \\=3f(8)+2f(7)+3f(6)+f(5)+2f(4) \dots$
Since $1$, $3$, and $5$ are all odd, there must be an even number of summands to get to $12$. You can't do it with just two summands, but you can with any even number from four to twelve:
$$\begin{align} \{3,3,3,3\}&\times1\\ \{5,3,3,1\}&\times12\\ \{5,5,1,1\}&\times6\\ \{3,3,3,1,1,1\}&\times20\\ \{5,3,1,1,1,1\}&\times30\\ \{3,3,1,1,1,1,1,1\}&\times28\\ \{5,1,1,1,1,1,1,1\}&\times8\\ \{3,1,1,1,1,1,1,1,1,1\}&\times10\\ \{1,1,1,1,1,1,1,1,1,1,1,1\}&\times1 \end{align}$$
where the number after the $\times$ is the number of different ways the summands in the given set can be arranged. This gives a total of
$$1+12+6+20+30+28+8+10+1=116$$
This is the quickest way I can think of to get the count for the specific number $12$. (If there's a quicker way, I'd like to see it!) In general, for larger numbers, the recursive formula
$$f(n)=f(n-1)+f(n-3)+f(n-5)$$
as explained in gandalf61's answer, is undoubtedly better. The sequence of $f(n)$'s is A060961 in the OEIS.
