Nt1330 Unit 4 Assignment

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On September 15, we did a group project where our goal was to figure out the equalities between the many objects that were given to us throughout the class. I am writing about the section in 7 through 9 that all had to do about the pebbles of different colors, the pebbles of the same color, and the M&M’s that dealt with all different colors. In section 7, we had to figure out the similarities of five bottles of different color M&M’s and group them up into sections that showed equality or that were different completely. My group and I separated the bottles into two sections. One section had two bottles that had yellow, orange, and brown M&M’s in them; the other section had the three bottles that had the different colors in jars like, one bottle had blue and …show more content…

This is the way to find the number of ways to choose “k” things from a set of “n”. Now for the stone situation we have above, we have “n” be the three different total colors you can have or want and we have “k” be the number of stones you want to choose from. We can pick up “k” stones (presumably in a row), so the number of ways to lay out the stones in a row is n! So every time we pick a stone, we have 1 less to choose from in “n”, so for the 1st stone we have “n” choices, for the 2nd stone we have (n-1) choices, and etc, and then eventually you multiply them together to find the number of arrangements. Now we have to choose from “k”. Obviously, there are “k” stones, but we need the number of ways those stones are arranged too, so k! However it doesn’t end there, cause you still have (n-k) stones remaining, and you need to arrange those in the same sets of the k!, therefore you need: k!*(n-k)! Since you’re looking for the number of “k” in “n”, you find the total number of arrangments and you divide out every single possible arrangments, to find the remaining ways to grab “k”, therefore that becomes: n!/(k!*(n-k)!). That’s

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