I have four positive numbers $a_1,\dots,a_4$, each less than $45$. How many different ways are there for $a_1+a_2+a_3+a_4<90$? I require different permutations i.e $a_1a_2a_3a_4$ is different from $a_2a_1a_4a_3$
Sum of four numbers less than a particular value
1
$\begingroup$
combinatorics
permutations
-
2Any numbers? Perhaps you mean integers, or positive integers? Your question is ambiguous. If any numbers are allowed, there are infinitely many combinations. – 2012-11-09
-
1Also, would you consider $a_1=21$, $a_2=a_3=a_4=20$ to be a different answer than $a_2=21,$ $a_1=a_3=a_4=20$, or the same thing? That makes a **crucial** difference. It looks like you probably interpret them to be different, given the `permutations` tag. Is this correct? – 2012-11-09
-
0it should be positive integers – 2012-11-09
-
0Rolled back: The previous edit introduced 3 changes: each a_i should be *positive*, each a_i is *less than* 45, the sum of the a_i's should be *less than* 90. – 2012-11-09
-
0Related: [Four positive numbers less than 10 must add up to 12](http://math.stackexchange.com/questions/372624/probability-of-random-integers-digits-summing-to-12). – 2013-08-15