I to be not too sure exactly how to execute this problem. Do I simply do $_14C_1$ (Single scoop) + $_13C_1$ (Double scoop)? Or is there a different way to handle this problem?
You have actually the solitary scoops correct, yet the double scoops would be $14choose2$. For this reason you have $14+14choose2$ possibilities.
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From an educational point of view, ns think it"s wise to autumn the Pr Cr, and look at simply the numbers because that a moment. Of course, there space 14 different single scoops.
A double scoop deserve to be make by adding a different scoop to a solitary scoop. For every solitary scoop, we have 13 different flavours, for this reason we finish up v $14 + 14 * 13 = 14^2$ different scoops in total.
Now a subtlety is the a cone v for instance vanilla ~ above the bottom and also chocolate on height is counted as different from a cone through vanilla on top and chocolate top top the bottom. The cones room what we (in usmam.org) speak to a permutation of every other: castle contain the exact same things and the exact same amount of them, just in a various order.
The question if cones through the same flavours yet different stacking have to be treated together the same is no addressed in this exercise. We currently counted them as if they were various (I expect you can understand why, this is difficult for me to define in words: a photo of a usmam.orgematics tree might assist here, however it"s a lot of effort to make one best now). All we have to do to correct for this, is divide the permutations for the twin scoops by two (since every double scoop mix has 2 permutations). Therefore we have $14 + frac14 * 132 = 14 + 7 * 13 = 105$ various combinations.
Of course, you can additionally use nCr nPr on your calculator, however I just wanted to present that the usmam.org behind it is pretty simple to understand and also you don"t really require a calculator for small examples favor this.
(The adhering to paragraph is not important to recognize my answer, yet hopefully the shines some light on combinations and permutations).
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Typically, if you have n different things, there room $n! = n * (n - 1) * (n - 2) * ... * 2 * 1$ different orders (so $n!$ different permutations). We have the right to derive this rather easily. We have the right to make every bespeak by first choosing the first element, climate the second, and so on, until we have actually picked every elements. Obviously, for the first element, we have n choices, because that the 2nd $n - 1$ (because we currently picked one facet that we for this reason can"t use again), because that the 3rd $n - 2$ (for the same reason), for the 3rd $n - 3$, and also so on. With the same logic, us don"t have to necessarily avoid until there room no much more elements left, we can additionally stop earlier.
Using this method, if we specify a nPr b as "How much various permutations space there if we select a sequence of b things from a set of a different things, without picking one twice?", we discover $a ext nPr b = a * (a - 1) * (a - 2) * ... * (a - (b - 1)) = a! / (a - b)!$. Notice that there room $b$ factors, which correspond to the $b$ times we have to pick the next element in the order. Now if we desire to count just the combinations and thus regard various permutations together the same if they have actually the same things, we just need to division by the number of permutations that $b$ things. This means dividing by $b!$. Therefore we have $a ext nCr b = a! / (b! * (a - b)!)$