Can integers be repeated?
i.e. can we have 2,2,5 ?
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?
4 answers
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.
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