How many subsets are there for this case?












1














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?






combinatorics permutations







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  • On this site, we use a type of $LaTeX$ called MathJax. Search for a tutorial on it. Please use it in future :)
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    Nov 25 at 4:10
















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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?






combinatorics permutations







share|cite|improve this question
























  • On this site, we use a type of $LaTeX$ called MathJax. Search for a tutorial on it. Please use it in future :)
    – Shaun
    Nov 25 at 4:10














1












1








1







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?






combinatorics permutations







share|cite|improve this question















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?






combinatorics permutations




combinatorics permutations






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edited Nov 25 at 4:14









Rócherz

2,7262721




2,7262721










asked Nov 25 at 4:06









Toby

1577




1577












  • On this site, we use a type of $LaTeX$ called MathJax. Search for a tutorial on it. Please use it in future :)
    – Shaun
    Nov 25 at 4:10


















  • On this site, we use a type of $LaTeX$ called MathJax. Search for a tutorial on it. Please use it in future :)
    – Shaun
    Nov 25 at 4:10
















On this site, we use a type of $LaTeX$ called MathJax. Search for a tutorial on it. Please use it in future :)
– Shaun
Nov 25 at 4:10




On this site, we use a type of $LaTeX$ called MathJax. Search for a tutorial on it. Please use it in future :)
– Shaun
Nov 25 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!






share|cite|improve this answer

















  • 1




    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.
    – Rócherz
    Nov 25 at 4:34










  • I think what you did makes a lot of sense to me! This should be right! :)
    – Toby
    Nov 25 at 4:37











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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!






share|cite|improve this answer

















  • 1




    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.
    – Rócherz
    Nov 25 at 4:34










  • I think what you did makes a lot of sense to me! This should be right! :)
    – Toby
    Nov 25 at 4:37
















1














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!






share|cite|improve this answer

















  • 1




    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.
    – Rócherz
    Nov 25 at 4:34










  • I think what you did makes a lot of sense to me! This should be right! :)
    – Toby
    Nov 25 at 4:37














1












1








1






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!






share|cite|improve this answer












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!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 at 4:17









mathnoob

1,775322




1,775322








  • 1




    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.
    – Rócherz
    Nov 25 at 4:34










  • I think what you did makes a lot of sense to me! This should be right! :)
    – Toby
    Nov 25 at 4:37














  • 1




    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.
    – Rócherz
    Nov 25 at 4:34










  • I think what you did makes a lot of sense to me! This should be right! :)
    – Toby
    Nov 25 at 4:37








1




1




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.
– Rócherz
Nov 25 at 4:34




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.
– Rócherz
Nov 25 at 4:34












I think what you did makes a lot of sense to me! This should be right! :)
– Toby
Nov 25 at 4:37




I think what you did makes a lot of sense to me! This should be right! :)
– Toby
Nov 25 at 4:37


















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