The answer is 7 R 4! hoped this helps!
Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:
![\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac2n%5Cright%5D%2C%5Cleft%5B%5Cdfrac2n%2C%5Cdfrac4n%5Cright%5D%2C%5Cleft%5B%5Cdfrac4n%2C%5Cdfrac6n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7B2%28n-1%29%7Dn%2C2%5Cright%5D)
Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

where
. Each interval has length
.
At these sampling points, the function takes on values of

We approximate the integral with the Riemann sum:

Recall that

so that the sum reduces to

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

Just to check:

Answer:
1200
Explanation:
Order does not matter, if we said xyz order, it would still not make a difference if it was zyx or yzx hence we use the combination formula:
nCr = n! / r! * (n - r)!
where n= total number of items
r= number of items chosen at a time
Combinations are used when the order of events do not matter in calculating the outcome.
We calculate using the formula:
(30×20×12)÷3!=1200
There are therefore 1200 ways for the three distinct items to not be in same row or column
D. x-6
That is the correct answer.