1. 6.4b The Piano Method (aka: Sticks and Stones Method)

2. Piano Method:
The piano method is for counting how many combinations that add to a certain value
One main concern is whether if any numbers can be zero or not
The numbers on the bottom indicate the sum
The black keys are where you can divide the values of each number
Let’s start with none of the numbers can be zero
How many ways can 3 numbers add to 10?
The sum is 10, so one divider is placed at the end
So you have 9 dividers, you can choose any 2 1 2 3 4 5 6 7 8 9 10

3. Sample Problem: Suppose “a”, “b”, and “c” are positive integers, and a+b+c=8, then how many different combinations of (a,b,c) are there?
Note: in this problem, a,b, & c can be the same
Also, all values are bigger than zero
Only restriction is that they add to 8
Solution: since the sum adds to 8, the black key in the end must be chosen
With 7 black keys left, we choose 2 to split the width into 3 numbers
Answer: 1 2 3 4 5 6 7 8

4. Practice: You have a blue, red, and green dice
Practice: You have a blue, red, and green dice. How many ways are there to roll a sum of 7 with three standard six sided dice? 1 2 3 4 5 6 7
There are 3 dice and none of them can be zero
The sum is 7, so one divider must be in the far right
You have 6 black keys and you can only choose 2 of them
Answer:

5. Sticks and Stones Challenge
Suppose you have four rocks and 2 sticks, how many ways can you place the two sticks between the rocks?
The sticks can be together or apart
There are 5 spots, you must choose 2

6. Now suppose you have 5 rocks and 3 sticks, how many ways can you place the 3 sticks?
First count how many ways when they are all together
Second, count when all 3 are apart
Third, count when it’s split between a pair and single

7. Sticks and Stones Method
One way to perceive this method is that each stick you put down is a permutation, the order matters
So your first stick will have 6 options
Since order matters, the next one can be in front or behind the 1st stick
The next stick will have 7 options!
The 3rd stick will have 8 options
Since order shouldn’t matter, you divide your answer by 3!

8. What if zeroes were included?
If zeroes are included, then you need to include the zero space at the left
One divider must be placed on the far right for the sum of the values
Each time you pick a divider, you increase the number of options for the next divider by one (left/right)
Treat the problem as a “choosing” question, “n” choose “r”
“n” will be the sum of “total objects” and “the number of dividers” minus 1
“r” will be the “number of dividers” minus 1
Minus 1 b/c one divider is placed for the sum 1 2 3 4 5 6 7 8 9 10

9. Ex: Suppose “a”,“b” , and “c” are whole numbers, how many different combinations of (a,b,c) are there if a+b+c=10
“a” “b” and “c” are whole numbers means they can be zero
If zeroes were included, we need to add another “black” key at the left, so the first number can be zero
There are 3 terms “a,b,c”. Sum is 10, so we get only 2 dividers
So instead, we use a permutation
11 options for the first key
Number of options for 2nd key increases by 1
You can put it on the left or right of the key chosen before {b/c you can have zero} 1 2 3 4 5 6 7 8 9 10

10. Ex: Suppose David has 12 dollars and he wants to give it to his 4 students. How many ways can he distribute his money?
Treat this question like: a+b+c+d=12, where a,b,c,&d can be equal to zero
You have 13 spots for four dividers
Since the sum must be 12, one of the dividers must be in the back, that leaves you with only 3 dividers left
First divider has 13 options
For each divider you pick, you increase by one more space (left/right)
This means the next divider has 14 options
The 4rd divider will then have 15 options
Since the 3 dividers don’t have order, divide by 3! 1 2 3 4 5 6 7 8 9 10 11 12

11. You want to buy a dozen donuts from Tim Horton’s and they only have 5 flavors left. How many ways can you purchase your donuts?
This question is like the previous one , where a+b+c+d+e=12, where a,b,c,d&e can be equal to zero
You have 13 spots for four dividers, one divider must be in the back to get a sum of 12
First divider has 13 options
Second divider has 14 options
3rd divider has 15 options, 4th divider has 16 options
Since the 4 dividers don’t have order, divide by 4! 1 2 3 4 5 6 7 8 9 10 11 12

12. Suppose you must order one of each flavor, how many ways can you purchase your dozen donuts?
This question is like the previous one , where a+b+c+d+e=12, but a,b,c,d&e can not be equal to zero
One divider must be in the back to get a sum of 12
You have 11 options for your 4 dividers
Note: if none of the numbers can be zero,
1. then get rid of the zero space in the left
2. Don’t increase by one after each turn 1 2 3 4 5 6 7 8 9 10 11 12

13. Challenges to Piano Method:
The question becomes more difficult when
1. None of the numbers can be the same
2. When we have inequalities
Suppose “a, “b” and “c” are positive integers, where 0 < a < b < c, how many different combinations of (a,b,c) are there such that:
i) a + b + c =10
Note: “a”, “b”, or “c” are not the same
“c” is the biggest, “a’ is the smallest
ii) a + b + c =20
The values a”, “b”, and “c” are not equal to zero
iii) a + b + c < 10
The last two are inequalities, so the sum can be less than or equal to 10
iv) a + b + c <100

14. Piano Method:
What would you do if none of the terms can be equal?
Ex: Suppose a + b + c = <a<b<c 1 2 3 4 5 6 7 8 9 10
With the 9 black keys, we need to select two of them to split the width of 10 into 3 separate parts
Since numbers are different, subtract all the triples & dbles
Permutation of every 6 are the same, divide by 6
We now have 24 combos left, but every 6 is a repeat
Triples:
(none)
Doubles:

15. There are 24 combinations
Suppose “a, “b” and “c” are positive integers, where 0 < a < b < c, how many different combinations of (a,b,c) are there such that: a + b + c = 20
1. There are 19 black keys, choose 2 keys to make 3 numbers that add to 20
2. Next we subtract all the triples and doubles because a,b,c must be different
3. After we subtract all the triples and doubles, divide by 6 because every 6 is a repeat
Triples:
(none)
Doubles:
There are 24 combinations

16. Piano method with inequalities
Sample question: a + b + c < 10
The sum is less than 10, can be 9, 8, 7, 6…
With the black keys, instead of choose 2, we choose 3 because the third key on the right determines the sum of the three numbers
Sum =6
Sum =9
Sum =10 1 2 3 4 5 6 7 8 9 10
The hardest part would be finding all the doubles and triples

17. 2. Count the triples and doubles and then subtract them
Suppose “a, “b” and “c” are positive integers, where 0 < a < b < c, how many different combinations of (a,b,c) are there such that: a + b + c < 10
1. Count the total number of combinations where order doesn’t matter and repeats are included
2. Count the triples and doubles and then subtract them
Triples:
3. After we subtract all the triples and doubles, divide by 6 because every 6 is a repeat
Doubles:
There are 7 combinations

18. There are 24952 combinations
Suppose “a, “b” and “c” are positive integers, where 0 < a < b < c, how many different combinations of (a,b,c) are there such that: a + b + c < 100
1. Count the total number of combinations where order doesn’t matter and repeats are included
2. Count the triples and doubles and then subtract them
Triples:
3. Subtract all the triples & doubles, divide by 6 because every 6 is a repeat
There are combinations
Doubles:

19. 1 2 3 4 5 Given that: How many different triples of are possible?
1. Treat this question like the piano method such that where you place the keys will be the numbers you get
2. Since there are 3 numbers, you place 3 arrows 1 2 3 4 5
3. Therefore, the first arrow has 6 options
4. The next arrow has 7 options
5. The next arrow has 8 options
6. Since the order doesn’t matter, we divide by 3!

20. There are 10 parking spaces in a row. 6 cars come in and park
There are 10 parking spaces in a row. 6 cars come in and park. A truck comes in afterwards and requires 2 adjacent spaces to park. What is the probability that the truck can park?
First Method:
1. Find the complement  How many ways can he NOT park the truck (none of the empty spaces are next to each other)
6 cars, 7 choices for the 4 empty spaces
2. Count the number of ways the 4 spaces can be distributed

21. There are 10 parking spaces in a row. 6 cars come in and park
There are 10 parking spaces in a row. 6 cars come in and park. A truck comes in afterwards and requires 2 adjacent spaces to park. What is the probability that the truck can park?
Second Method:
1. Place the four empty parking spaces down
2. Put 3 cars in between these empty parking spots so they can’t be next to each other
3. Now we look at the ways to place the remaining 3 cars
4. First car has 5 options
5. Second car has 6 options
6. 3rd car has 7 options
7. Since order doesn’t matter, divide by 3!

22. There are 4 girls and 10 guys standing in a line
There are 4 girls and 10 guys standing in a line. In how many ways can they be arranged if none of the girls are standing next to each other?
1. This question is the same as the previous one, put the 4 girls down first and in between them, place a guy
2. Three boys must be placed in between to separate them, so 7 boys left to be placed where ever
3. With the 7 remaining boys, 1st one has 5 spots, next has 6, …..

23. If you have 3 black socks and 6 gold socks, how many ways can you place them so that there will always be 2 gold socks between any two black
4 socks are placed, 2 are left
If you have 5 black socks and 10 gold socks, how many ways can you place them so that there will always be 2 gold socks between any two black
8 socks are placed, 2 are left
With “n” black socks 