Register to reply 
Combinatorics: number of words 
Share this thread: 
#1
Mar1909, 04:47 PM

P: 254

ere me now.
I'm trying to figure out how many words of length r having exactly k distinct letters can be made with an alphabet of size n. Call this number a_k,r. It satisfies the recursion [tex]a_{k,r} = ka_{k,r1} + (nk+1)a_{k1,r1}[/tex] [tex]a_{1,r} = n[/tex] for r >= 1 [tex]a_{k,r} = 0[/tex] for r < k My best effort so far is to use a generating function as follows [tex]f_k(x) = \sum_{r=1}^{\infty}a_{k,r}x^r[/tex] and use the recursion formula to get [tex]f_k(x) = \frac{(nk+1)x}{1kx}f_{k1}(x) = \frac{x^{k1}(n1)(n2)...(nk+1)} {(1x)(12x)...(1kx)}f_1(x)[/tex] or since f_1(x) = n/(1x), [tex]f_k(x) = \frac{(k1)!\left(^n_{k1} \right)x^{k1}}{(1x)(12x)...(1kx)}[/tex] I'm having a well difficult time getting the coefficient of x^r (which would be a_k,r) in the expansion of this. Is there a way to simplify it so you don't have k nested sums? thanks. 


#2
Mar1909, 05:56 PM

P: 34

I am trying to follow your approach and the f's confuse me: the problem I have is that
[tex] f_{k}(1) = \sum_{r=1}^{\infty} a_{k,r} \geq 1, [/tex] but by your recursion relation we have [tex] \frac{f_{k}(1)}{f_{k1}(1)} = \frac{nk+1}{1k}. [/tex] The right hand side must be positive while the left must be negative. This means either I have misunderstood something or you have... I am sure some combinatorist has got a simply answer for this... but pending them stating what it is here is a way to at least reduce the problem. First find how many ways to pick k distinct letters from the alphabet with n letters (order of picking not important). Consider these letters as a subalphabet. From this subalphabet you will need to create words that are length r and use every letter. If you find this number then you can multiply by the previous to obtain the answer. Please tell me if you find a simple answer and\or which one of us made the mistake. 


#3
Mar2009, 08:15 PM

P: 254

Yes, the series expansion for the f_k's is only valid for x < 1/k. But
your suggested way of doing it is a lot better. For an alphabet of k letters, the number b_k,r of rletter words which have all k letters is k^r minus the number of words in which at least one of the letters is absent. I calculated the latter using the sieve principle as # of rletter words missing at least 1 of the k letters = [tex]\left( ^{k}_{1} \right)(k1)^r  \left( ^{k}_{2} \right)(k2)^r +...+ (1)^{k}\left( ^{k}_{k1} \right) = \sum_{j=1}^{k1}(1)^{j1}\left( ^{k}_{j} \right)(kj)^r[/tex] Then subtracting this from k^r gives [tex]b_{k,r} =\sum_{j=0}^{k1}(1)^{j}\left( ^{k}_{j} \right)(kj)^r [/tex] As you said, to generalize this to n > k letters in the alphabet, multiply b_k,r by choose(n,k) to get [tex]a_{k,r} = \left( ^{n}_{k} \right)b_{k,r}[/tex] For the special case k = r, this should simplify to [tex]a_{r,r} = r!\left( ^{n}_{r} \right)[/tex] I've been trying to verify by induction that [tex]\sum_{j=0}^{r1}(1)^{j}\left( ^{r}_{j} \right)(rj)^r = r![/tex] but I can't seem to do it. It seems to be true though, if you plug in small numbers for r. Thanks for the tip! At least I'm making progress now. If you know of a way to simplify the above answer further, please show it. 


#4
Mar2109, 06:08 AM

P: 10

Combinatorics: number of words
[tex]\sum_{j=0}^{r1}(1)^{j}\\( ^{r}_{j} \\)(rj)^r = r![/tex]
This expression is easy to prove by considering the rth derivative of (1e^x)^r at x=0 


#5
Mar2109, 11:01 PM

P: 254




Register to reply 
Related Discussions  
Simple Combinatorics: What are odds of picking same number?  General Math  5  
Combinatorics  Precalculus Mathematics Homework  5  
Combinatorics  Calculus & Beyond Homework  7  
Combinatorics problem  Introductory Physics Homework  7  
In how many ways to answer a truefalse test with 6 questions  Introductory Physics Homework  3 