How many subsets are there for this case?
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
edited Nov 25 at 4:14
Rócherz
2,7262721
asked Nov 25 at 4:06
Toby
1577
add a comment |
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
edited Nov 25 at 4:14
Rócherz
2,7262721
asked Nov 25 at 4:06
Toby
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
add a comment |
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
edited Nov 25 at 4:14
Rócherz
2,7262721
asked Nov 25 at 4:06
Toby
1577
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
edited Nov 25 at 4:14
Rócherz
2,7262721
asked Nov 25 at 4:06
Toby
1577
edited Nov 25 at 4:14
Rócherz
2,7262721
asked Nov 25 at 4:06
Toby
1577
edited Nov 25 at 4:14
Rócherz
2,7262721
edited Nov 25 at 4:14
Rócherz
2,7262721
edited Nov 25 at 4:14
Rócherz
2,7262721
2,7262721
asked Nov 25 at 4:06
Toby
1577
asked Nov 25 at 4:06
Toby
1577
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
add a comment |
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
add a comment |
1 Answer
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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!
answered Nov 25 at 4:17
mathnoob
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
add a comment |
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1 Answer
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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!
answered Nov 25 at 4:17
mathnoob
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
add a comment |
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!
answered Nov 25 at 4:17
mathnoob
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
add a comment |
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!
answered Nov 25 at 4:17
mathnoob
1,763422
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!
answered Nov 25 at 4:17
mathnoob
1,763422
answered Nov 25 at 4:17
mathnoob
1,763422
answered Nov 25 at 4:17
mathnoob
1,763422
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
add a comment |
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
add a comment |
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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 :)
– Shaun
Nov 25 at 4:10