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.

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 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 &#34;tree ofpossibilities.&#34; 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 &#34;root&#34; of the tree.Attached to the root are five “branches,&#34; 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 &#34;leaves&#34; 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 &#34;words&#34; in alphabetic or lexicographic order.

Accepted Answer

This is a basic recursion problem where we need to find all the sequences of set {1,3,5,7,9} and…