A local kindergarten is thinking of making posters that show all the different ways of adding two or more integers from 1 to 9 to get a sum of 10. If there is enough space on each poster for up to 50 possible solutions, how many posters will the school need to make?
(Note: sums that contain the same number but in a different order are considered to be different; for example, 1 + 9 and 9 + 1 are two different solutions.)
What is the answer to this problem, and more importantly, how do I solve it?
Can integers be repeated?
i.e. can we have 2,2,5 ?
yup
two numbers:
1-9, 2-8, 3-7, 4-6, 5-5, ... 9 of them
three numbers:
1-1-8, arrange in 3 ways
1-2-7, arrange in 6 ways
1-3-6, arrange in 6 ways
1-4-5, arrange in 6 ways
2,2,6, arrange in 3 ways
2,3,5, arrange in 6 ways
2,4,4, arrange in 3 ways
3,3,4, arrange in 3 ways .... 36 of those
four numbers:
1,1,1,7 , arrange in 4 ways
1,1,2,6, arrange in 12 ways
1,1,3,5 , arrange in 12 ways
1,1,4,4, arrange in 6 ways
... There will be 84 of these
five numbers:
e.g. 2,2,2,1,3 arranged in 20 ways
etc
.... there will be 126 of those
6 numbers:
e.g. 1,1,1,2,2,3 arrange in 60 ways
etc
.... there will be another 126 of those
Notice I am getting the elements of row 10 of Pascal's triangle, so
the total will be 9+36+84+126+126+84+36+9+1 = 511 or (2^9 - 1)
(missing the 1 at the left since we wanted the sum of 2 or more)
since each poster contains 50 entries , you will need 511/50 or 11 posters.
Thx! that the answer I got
To solve this problem, we need to find all possible ways to add two or more integers from 1 to 9 in order to get a sum of 10. We also need to consider that sums with the same numbers but in a different order are considered different solutions.
To tackle this problem, we can start by listing all the possible combinations of numbers that add up to 10. Since the integers range from 1 to 9, we can have:
1 + 9
2 + 8
3 + 7
4 + 6
5 + 5
6 + 4
7 + 3
8 + 2
9 + 1
Now, let's look at the combinations that involve more than two integers:
1 + 1 + 8
1 + 1 + 1 + 7
1 + 1 + 1 + 1 + 6
1 + 1 + 1 + 1 + 1 + 5
1 + 1 + 1 + 1 + 1 + 1 + 4
1 + 1 + 1 + 1 + 1 + 1 + 1 + 3
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 2
(Note: There is only one combination that involves three integers, followed by increasing combinations involving more 1's and less of the other integers in each subsequent combination.)
We can observe a pattern here in terms of the number of 1's included in each combination. As we move from left to right, the number of 1's increases by one, and the other integers decrease by one.
To determine the number of combinations involving more than two integers, we can calculate the number of ways to arrange these 1's and 2's. Since we have two types of digits, let's use a variation of the "Stars and Bars" method, known as "Balls and Urns" or "Stars and Bars with restrictions." In this case, the stars represent the 1's, and the bars represent the 2's.
With 1's and 2's, we have three positions: before the first bar, between the first and second bar, and after the second bar. Since we have nine 1's and two 2's, we need to distribute them among these positions.
Using a combinations formula, the number of ways to arrange these elements is given by:
C(9 + 2 - 1, 9) = C(10, 9) = 10
So, there are 10 combinations involving more than two integers.
Taking into account the combinations involving two integers, which we listed earlier, we have a total of 9 + 10 = 19 different combinations.
Since each poster can hold up to 50 solutions, we need:
19 combinations / 50 solutions per poster = 0.38
Since we can't have a fraction of a poster, the school will need to make at least 1 poster to accommodate all the solutions.
Therefore, the answer to this problem is 1 poster.
To solve this problem, you need to list all possible combinations and analyze the patterns within the combinations. Furthermore, understanding how to apply the "Stars and Bars" method can help identify the number of combinations involving more than two integers. Finally, performing the necessary calculations will give you the final answer.
Note: The above explanation assumes that the school needs to show all unique combinations of integers that sum up to 10. If the question asks for the total number of unique combinations without listing them individually, the calculation can be simplified.