# Sequences from the positive integer numbers $(a,b,c,d,e)$

How to find how many sequences from the positive integer numbers $(a,b,c,d,e)$,such that :

$$abcde \le a+b+c+d+e \le 10$$

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I would try to solve this logically. If 10 is the maximum number of a + .... + e, then that means the numbers must be limited to 6. Now let's find out which sets of numbers excluding 1 gives products less than 10:
2,2
2,3
2,4 || 2,2,2
2,5
3,3

Any other sets of numbers multiplied will give values greater than 10 for abcde. So you can try (1,1,1,2,2), (1,1,1,2,3), (1,1,1,2,4), (1,1,1,2,5), (1,1,1,3,3), and lastly (1,1,2,2,2). We can't forget single numbers by themselves either. (1,1,1,1,1), (1,1,1,1,2), (1,1,1,1,3), (1,1,1,1,4), (1,1,1,1,5), and (1,1,1,1,6). Now make sure the sums are all greater than the product.
1,1,1,2,2:4:7-good
1,1,1,2,3:6:8-good
1,1,1,2,4:8:9-good
1,1,1,2,5:10:10-good
1,1,1,3,3:9:9-good
1,1,2,2,2:8:8-good
and obviously, all of the (1,1,1,1,x) sets also work.
So now we have to make combinations of each type of set, since (1,1,1,1,2) is different than (1,1,1,2,1).
For sets in the form of (a,a,a,a,a), there are 5!/5! = 1 combinations. For sets in the form of (a,a,a,a,b), there are 5!/4! = 5 combinations.
For sets in the form of (a,a,a,b,b), there are 5!/(3!*2!) = 10 combinations.
For sets in the form of (a,a,a,b,c), there are 5!/3! = 20 combinations.
There are 1 set of the first type, 5 sets of the second type, 3 sets of the third type, and 3 sets of the last type.
This leaves us with 1*1 + 5*5 + 3*10 + 3*20 = 1 + 25 + 30 + 60 = 116.

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WAIT..hold up i am checking my work...I made a java program but its giving me 76...Let me check my work sorry. –  mathguy Aug 9 '12 at 3:00
Nevermind made an error in the program. 116 is correct! –  mathguy Aug 9 '12 at 3:03
A hand count using slightly different idea (partitions) gives same $116$. –  André Nicolas Aug 9 '12 at 3:06