If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct? In other words, how many 5-tuples of integers (h, i, j, k, m) are there with 1 ≤ h ≤ i ≤ j ≤ k ≤ m ≤ n? As in Example 9.6.3, you can represent any ordered 5-tuple of integers (h, i, j, k, m) with 1 ≤ h ≤ i ≤ j ≤ k ≤ m ≤ n as a string of n − 1 vertical bars and 5 crosses, with the position of crosses indicating which 5 integers from 1 to n are included in the 5-tuple. Thus, the number of 5-tuples is the same as the number of strings of n+4 vertical bars and 5 crosses, which is n(n+1)(n+2)(n+3)(n+4) 120​ .

Step-by-step explanation:

Given

5 tuples implies that:

implies that:

Required

How many 5-tuples of integers are there such that

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:

Expand the numerator

Solved

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step-by-step explanation:

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204