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?










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














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?










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?










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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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










1 Answer
1






active

oldest

votes


















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012429%2fhow-many-subsets-are-there-for-this-case%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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














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

1,763422




1,763422








  • 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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012429%2fhow-many-subsets-are-there-for-this-case%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Ellipse (mathématiques)

Quarter-circle Tiles

Mont Emei