There are 4 components (k1, k2, k3, k4). a) How many sequences are there of length 1000?
I know that the number of strings of length n with k elements is k^n, so 4^1000.
b) There is the same amount of k1 as there is of k3 and the same amount of k2 as k4, but there is 3 times the amount of k2 as k1. How many such sequences are there of length 1000?
I'm so lost on how to do this problem.
On
If there are $m$ k2s, then there are $m$ k4s and $3m$ k1s and k3s.
The total number of items is $3m+m+3m+m = 8m =1000$, so $m=125$.
The k2s can be in $\binom{1000}{125} $ positions. The k1s can be in $\binom{1000-125}{3\cdot 125} =\binom{875}{375} $ positions. The k3s can be in $\binom{875-375}{3\cdot 125} =\binom{500}{375} $ positions. The k4s are in the remaining positions.
The total number of possibilities is
$\begin{array}\\ \binom{1000}{125}\binom{875}{375}\binom{500}{375} &=\dfrac{1000!}{125!875!}\dfrac{875!}{375!500!}\dfrac{500!}{375!125!}\\ &=\dfrac{1000!}{125!^2375!^2} \end{array} $
You know that every letter is either $k_1,k_2,k_3$, or $k_4$. If $a_1,a_2,a_3$, and $a_4$ denote the number of times the respective letters are in your sequence, this means $a_1+a_2+a_3+a_4=1000$. But now we additionally know that $a_1=a_3$, $a_2=a_4$, and $3a_1=a_2$. Plugging this in gives $$ 8a_1=1000 $$ or $a_1=125$. So we know there are $125$ copies of $k_1$ and $k_3$, and $375$ copies of $k_2$ and $k_4$. Of your $1000$ letters, first choose the $125$ locations where $k_1$ will be, then out of the remaining $875$ choose $125$ more for the $k_3$, and then finally choose where the $k_2$ will go; the remaining slots are forced to be filled with $k_4$.