Hello, in problem 2 of the combinatorics exercises, as part of the solution, it is written: In this case, it is vital to not only know which banners we are using, but also to know how many times we are using each one, so we can assign them accordingly. If instead, we did not care how many times we use each banner we have selected, then we would have to find the sum of C58+ C48+ C38+ C28+ C18. That is because we are essentially estimating the number of ways, we can select the banners, assuming we are using 5 different ones, 4 different ones, 3 different ones and so on this equates to 37448, but I am struggling to differentiate or understand how and why we need to do this and how it differs from variations with repetition equalling 32768? in my mind, whether we use 1 banner 2,3 or 4 times and another banner then 2,3 or 4 times, or if we only used 1 of the banners 5 times, this is the same as being a repeating value, is it not? the explanation suggests there are a further 4680 variations of the 8 banners across the 5 campaigns over and above the calculation of variations with repetition, but what factor causes the additional variation to just simply rotating all 8 banners through 5 places with repetition? Any insight or additional clarity, would be appreciated. Many thanks

Hey Daniel, I think you're using the wrong formula here based on "this equates to 37448". Since we don't care how many times we're using each banner, then the answer will certainly be less than 792. A quick computation lead me to the following results, 56 + 70 + 56 + 28 + 8 = 218. Best, 365 Vik

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Daniel Sweeting

Last answered:

15 Jun 2020

Posted on:

10 Jun 2020

problem 2 of the combinatorics exercises

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Viktor Mehandzhiyski

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Posted on:

15 Jun 2020

problem 2 of the combinatorics exercises

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