Let A be the set {1,3,5,7,9}. How many sequences on A sum to 6? 2.3.7 Counting Algorithmically (Without a Formula)In this subsection, we want to provide a strategy for counting sequences, using treediagrams, which can be applied when the set of possible values for s; depends onwhat the previous values in S actually are. We'll look at two examples.Example 2.3.6: Sequences on P that sum to 5How many sequences of positive integers have a sum that equals 5?// Both (1,4) and (1,1,2,1) sum to 5, so we'll have to deal with sequences of// varying lengths. But, since each entry is at least one, there can be at most 5// entries.Let's generate all such sequences, one entry at a time, by growing a "tree ofpossibilities." Suppose S= (s1, S2,. .., Sk) is such a sequence. The possible values forS1 are 1,2,3,4, and 5. This may be displayed bystart345The vertex labeled start, at the top of the diagram, is called the "root" of the tree.Attached to the root are five “branches," one for each possible value of s1. Thesepossible values of s1 are written inside the vertices (circles) at the ends of the 5branches.// Yes, the tree is upside down.If s1 = 1, then the possible values for s2 are 1,2,3, and 4, and we may add 4 more(downward) branches from the vertex corresponding to s1 = 1. If s1 = 2, then thepossible values for s2 are 1,2, and 3, and we may add 3 more branches from thevertex corresponding to s1 = 2. But if s1 = 5, then the current subsequence cannotbe extended; we cannot add more branches from the vertex corresponding to s1 = 5.These terminal vertices are called "leaves" and are indicated in the diagram assquares (rather than circles).From each vertex indicating a possible value of s2 that's not a leaf, we may addbranches corresponding to values of s3. And so on until we have a diagram of allsequences on P that sum to 5. start5421Now, by counting the leaves, we have determined that there are exactly 16sequences on P which sum to 5.We have consistently labeled the values below a vertex from smallest to largestacross the diagram from left to right. If we take the leaves as they occur across thediagram from left to right, and write down the corresponding sequence, we get1. (1,1,1,1,1)2. (1,1,1,2)// aaaaa// aaab// aaba3. (1,1,2,1)4. (1,1,3)// aac5. (1,2,1,1)// abaa6. (1,2,2)// abb7. (1,3,1)// aca8. (1,4)// ad// baaa// bab9. (2,1,1,1)10. (2,1,2)11. (2,2,1)// bba12. (2,3)// bc13. (3,1,1)// caa14. (3,2)// cb15. (4,1)// da16. (5)// e// There is something special about the order in which the sequences appear.// If the first 5 digits were replaced by the first 5 letters, the sequences of digits// produce "words" in alphabetic or lexicographic order.

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Question

Let A be the set {1,3,5,7,9}. How many sequences on A sum to 6?

2.3.7 Counting Algorithmically (Without a Formula)
In this subsection, we want to provide a strategy for counting sequences, using tree
diagrams, which can be applied when the set of possible values for s; depends on
what the previous values in S actually are. We'll look at two examples.
Example 2.3.6: Sequences on P that sum to 5
How many sequences of positive integers have a sum that equals 5?
// Both (1,4) and (1,1,2,1) sum to 5, so we'll have to deal with sequences of
// varying lengths. But, since each entry is at least one, there can be at most 5
// entries.
Let's generate all such sequences, one entry at a time, by growing a "tree of
possibilities." Suppose S= (s1, S2,. .., Sk) is such a sequence. The possible values for
S1 are 1,2,3,4, and 5. This may be displayed by
start
3
4
5
The vertex labeled start, at the top of the diagram, is called the "root" of the tree.
Attached to the root are five “branches," one for each possible value of s1. These
possible values of s1 are written inside the vertices (circles) at the ends of the 5
branches.
// Yes, the tree is upside down.
If s1 = 1, then the possible values for s2 are 1,2,3, and 4, and we may add 4 more
(downward) branches from the vertex corresponding to s1 = 1. If s1 = 2, then the
possible values for s2 are 1,2, and 3, and we may add 3 more branches from the
vertex corresponding to s1 = 2. But if s1 = 5, then the current subsequence cannot
be extended; we cannot add more branches from the vertex corresponding to s1 = 5.
These terminal vertices are called "leaves" and are indicated in the diagram as
squares (rather than circles).
From each vertex indicating a possible value of s2 that's not a leaf, we may add
branches corresponding to values of s3. And so on until we have a diagram of all
sequences on P that sum to 5.

start
5
4
2
1
Now, by counting the leaves, we have determined that there are exactly 16
sequences on P which sum to 5.
We have consistently labeled the values below a vertex from smallest to largest
across the diagram from left to right. If we take the leaves as they occur across the
diagram from left to right, and write down the corresponding sequence, we get
1. (1,1,1,1,1)
2. (1,1,1,2)
// aaaaa
// aaab
// aaba
3. (1,1,2,1)
4. (1,1,3)
// aac
5. (1,2,1,1)
// abaa
6. (1,2,2)
// abb
7. (1,3,1)
// aca
8. (1,4)
// ad
// baaa
// bab
9. (2,1,1,1)
10. (2,1,2)
11. (2,2,1)
// bba
12. (2,3)
// bc
13. (3,1,1)
// caa
14. (3,2)
// cb
15. (4,1)
// da
16. (5)
// e
// There is something special about the order in which the sequences appear.
// If the first 5 digits were replaced by the first 5 letters, the sequences of digits
// produce "words" in alphabetic or lexicographic order.

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