In this item, we are informed that the order of the entries does not matter in determination the number of ways in which the Archie can choose for his party. Because the arrangement or order is not important, this type of problem uses the concept of COMBINATION.
The equation for combination is,
nCr = n!/((n - r)!(r!))
nCr is read as "combination of n taken r".
Substituting the known values to the equation,
15C6 = 15! / ((15 - 6)!(6!))
= 5005
Hence, there are 5005 ways in which Archee can choose the 6 entrees for his party.
Answer:
I believe the answer is 24
Check the picture below.
so we know the radius of the semicircle is 2 and the rectangle below it is really a 4x4 square, so let's just get their separate areas and add them up.
![\stackrel{\textit{area of the semicircle}}{\cfrac{1}{2}\pi r^2}\implies \cfrac{1}{2}(\stackrel{\pi }{3.14})(2)^2\implies 3.14\cdot 2\implies 6.28 \\\\\\ \stackrel{\textit{area of the square}}{(4)(4)}\implies 16 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{sum of both areas}}{16+6.28=22.28}~\hfill](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20semicircle%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28%5Cstackrel%7B%5Cpi%20%7D%7B3.14%7D%29%282%29%5E2%5Cimplies%203.14%5Ccdot%202%5Cimplies%206.28%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20square%7D%7D%7B%284%29%284%29%7D%5Cimplies%2016%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20both%20areas%7D%7D%7B16%2B6.28%3D22.28%7D~%5Chfill)
Answer:
The factors of 8 are 1,2,4, and 8.
Step-by-step explanation: