Let n1<n2<n3<n4<n5 be positive integers such that n1+n2+n3+n4+n5=20 . Then the number of such distinct arrangements (n1, n2, n3, n4, n5) is _________

This problem can be solved using the concept of stars and bars in combinatorics.

Step 1: Since all the numbers are positive integers and distinct, let's assign each number a value of 1 to start with. This ensures that they are all positive and distinct. So, we have n1=1, n2=2, n3=3, n4=4, n5=5. This sums up to 15.

Step 2: We now have 20 - 15 = 5 left to distribute among the 5 numbers.

Step 3: We can think of this as placing 5 identical balls (the remaining sum) into 5 distinct boxes (the numbers n1 to n5).

Step 4: This is a standard stars and bars problem, and the solution is given by the binomial coefficient C(n + k - 1, k - 1), where n is the number of balls and k is the number of boxes.

Step 5: Substituting n = 5 (balls) and k = 5 (boxes), we get C(5 + 5 - 1, 5 - 1) = C(9, 4).

Step 6: The binomial coefficient C(9, 4) = 9! / [4!(9-4)!] = 126.

So, there are 126 distinct arrangements of (n1, n2, n3, n4, n5) that satisfy the given conditions.