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# 1. Some binomial coefficients

Prove that, for all non-negative integers n and k,

## 1.1. Solution

There are several ways to prove this. The easiest is perhaps by induction on n. First observe that

since either k = 0 and both binomial coefficients are 1 or k > 0 and both are 0.

For larger n, we have

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(Here the second-to-last step uses the induction hypothesis and the last uses Pascal's identity.)

# 2. More binomial coefficients

Prove that, for all non-negative integers n, m, and k,

## 2.1. Solution

Observe that the left-hand side counts the number of ways, starting with a set S of n elements, to choose a subset A of S with k elements and then choose a further subset B of A with m elements. On the right-hand side, we first choose B (a subset of S with m elements) and then choose the remaining k-m elements of A from the remaining n-m elements of S. Since both methods give us the same chain of subsets A, B, and S, we have a combinatorial proof of the identity.

(It's also possible to do this directly by expanding the binomial coefficients into factorials, but it's messy.)

# 3. Still more binomial coefficients

Give a simple expression for

## 3.1. Solution

This is a job for generating functions!

Compute:

# 4. 1, 2, 3

Let

• T(0) = 1, T(1) = 2, and

T(n) = 2T(n-1) + 3T(n-2) when n > 1.

Determine a simple non-recursive formula for T(n).

## 4.1. Solution

Do the generating-function trick of multiplying by zn and adding for all n:

Solve for F(z) to get

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and finally we can read off the coefficients T(n) = (3/4)*3n + (1/4)*(-1)n.

# 5. Forbidden substrings

A substring of a sequence x1x2...xn is a consecutive sequence of values xixi+1...xi+k that appears in the original sequence. So for example 010 is a substring of 00100 but not of 01100. An n-bit string is a sequence of n bits, where a bit is either 0 or 1.

Give the simplest expression you can as a function of n for

1. The number of n-bit strings that do not contain 0 as a substring.
2. The number of n-bit strings that do not contain 01 as a substring.
3. The number of n-bit strings that do not contain 00 as a substring.

## 5.1. Solution

1. 1 (the all-1's string).
2. n+1. Any string that doesn't contain 01 must be of the form 1k0n-k, since after the first 0 we can only have more 0's. The number of 1's can range from 0 to n, giving n+1 possibilities.

3. Let's partition the set of strings that don't contain 00 into those that are empty, those that start with 0 and those that start with 1. Let Ti(n) be the number of length-n strings that start with i. Then T(0) = 1 and T(n) = T0(n) + T1(n) when n > 0. But T0(n) = T1(n-1) for n > 1 since after the initial 0 we must continue with a 1 followed by a string that contains no 00's; for n = 1 we have the special case T0(1) = 1. For T1 we have T1(n) = T(n-1) since we can continue with either 0 or 1. We also have T0(0) = T1(0) = 0 since a zero-length string doesn't start with either 0 or 1.

Let's build up a table of values and see if we recognize the sequence:

 n T0(n) T1(n) T(n) 0 0 0 1 1 1 1 2 2 1 2 3 3 2 3 5 4 3 5 8 5 5 8 13

By now the pattern is pretty well established: for large n, T(n) = T0(n) + T1(n) = T1(n-1) + T(n-1) = T(n-2) + T(n-1), and the numbers in the last column look suspiciously like the Fibonacci_numbers Fib(n+2). To prove this, observe that T(0) = 1 = Fib(2) and T(1) = 2 = Fib(3), and for larger n the recurrence holds as discussed above.

We could stop here, arguing that Fib(n+2) is a pretty well-known sequence with a simple formula, or we could try to derive one using generating functions. (Looking in BiggsBook doesn't help us, although it is a common enough example that we could probably find the formula on the web somewhere.)

So if we have to build a generating function, perhaps we should do so directly. Summing up our various recurrences we have

From this solve for F to get F(z) = (1+z)/(1-z-z2). Extracting coefficients is the usual exercise in partial fraction expansion (which we'll omit in these solutions but which you should know how to do).

2014-06-17 11:57