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Question: You are given $20$ sugar bars (B$_{1}$, B$_{2}$, ..., B$_{20}$) and $50$ salt bars (S$_{1}$, S$_{2}$, ..., S$_{50}$). Consider subsets of these $70$ bars, that contain at least $3$ sugar bars (and any number of salt bars). How many such subsets are there?

(Answer: $1.18 \times 10^{21}$)

Attempt: Since it is at least $3$ sugar bars, it should start the count at $\dbinom{20}{3}$ all the way to $\dbinom{20}{20}$. For the salt bars, since it is any number of combinations, it should be $2^{50}$. I just multiplied them out using product rule but I didn't get the correct answer. How should I proceed with this?

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  • $\begingroup$ On this site, we use a type of $\LaTeX$ called MathJax. Search for a tutorial on it. Please use it in future :) $\endgroup$ – Shaun Nov 25 '18 at 4:10
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Total number of subsets is $2^{70}$. consider those that have less than 3 sugar bars, $\binom{20}{2}2^{50}+\binom{20}{1}2^{50}+2^{50}$. so we should get $2^{70}-\binom{20}{2}2^{50}-\binom{20}{1}2^{50}-2^{50}$. I hope I got this right!

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    $\begingroup$ It's hard to tell precisely how close to being right you are: \begin{align} 2^{70} -\dbinom{20}{2} 2^{50} -\dbinom{20}{1} 2^{50} -2^{50} &=1.1804 \times 10^{21}; \\ 2^{70} -\dbinom{20}{2} 2^{50} &=1.1804 \times 10^{21}; \\ 2^{70} -2^{50} &=1.1806 \times 10^{21}; \\ 2^{70} &=1.1806 \times 10^{21}; \end{align} But your reasoning seems legit. $\endgroup$ – Rócherz Nov 25 '18 at 4:34
  • $\begingroup$ I think what you did makes a lot of sense to me! This should be right! :) $\endgroup$ – Toby Nov 25 '18 at 4:37

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