Answer:
[tex]\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}[/tex]
Step-by-step explanation:
Given
5 tuples implies that:
[tex]n = 5[/tex]
[tex](h,i,j,k,m)[/tex] implies that:
[tex]r = 5[/tex]
Required
How many 5-tuples of integers [tex](h, i, j, k,m)[/tex] are there such that[tex]n\ge h\ge i\ge j\ge k\ge m\ge 1[/tex]
From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.
Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location
So, there are n + 4 items to make selection from
The selection becomes:
[tex]^{n}C_r => ^{n + 4}C_5[/tex]
[tex]^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}[/tex]
[tex]^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}[/tex]
Expand the numerator
[tex]^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}[/tex]
[tex]^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}[/tex]
[tex]^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}[/tex]
[tex]^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}[/tex]
Solved