Two basic combinational methods questions
up vote
1
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favorite
I am an absolute beginner in math. I have 2 questions about combinatorial methods, with their answers. I spent lots of time to understand these questions and tried to find a connection how the author found this solution, but I am literally lost. Can anybody explain me very basically, how the author solved this questions?
probability combinatorics combinations
edited Nov 18 at 10:42
N. F. Taussig
42.8k93254
asked Nov 8 at 10:05
Ozan
82
add a comment |
up vote
1
down vote
favorite
I am an absolute beginner in math. I have 2 questions about combinatorial methods, with their answers. I spent lots of time to understand these questions and tried to find a connection how the author found this solution, but I am literally lost. Can anybody explain me very basically, how the author solved this questions?
probability combinatorics combinations
edited Nov 18 at 10:42
N. F. Taussig
42.8k93254
asked Nov 8 at 10:05
Ozan
82
3
Please try writing the question out using mathjax math.meta.stackexchange.com/questions/5020/…. Also only enter one question at a time.
– Prakhar Nagpal
Nov 8 at 10:09
Are you aware of the combination/selection formula?
– Prakhar Nagpal
Nov 8 at 10:17
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am an absolute beginner in math. I have 2 questions about combinatorial methods, with their answers. I spent lots of time to understand these questions and tried to find a connection how the author found this solution, but I am literally lost. Can anybody explain me very basically, how the author solved this questions?
probability combinatorics combinations
edited Nov 18 at 10:42
N. F. Taussig
42.8k93254
asked Nov 8 at 10:05
Ozan
82
I am an absolute beginner in math. I have 2 questions about combinatorial methods, with their answers. I spent lots of time to understand these questions and tried to find a connection how the author found this solution, but I am literally lost. Can anybody explain me very basically, how the author solved this questions?
probability combinatorics combinations
probability combinatorics combinations
edited Nov 18 at 10:42
N. F. Taussig
42.8k93254
asked Nov 8 at 10:05
Ozan
82
edited Nov 18 at 10:42
N. F. Taussig
42.8k93254
asked Nov 8 at 10:05
Ozan
82
edited Nov 18 at 10:42
N. F. Taussig
42.8k93254
edited Nov 18 at 10:42
N. F. Taussig
42.8k93254
edited Nov 18 at 10:42
N. F. Taussig
42.8k93254
42.8k93254
asked Nov 8 at 10:05
Ozan
82
asked Nov 8 at 10:05
Ozan
82
asked Nov 8 at 10:05
Ozan
82
82
3
Please try writing the question out using mathjax math.meta.stackexchange.com/questions/5020/…. Also only enter one question at a time.
– Prakhar Nagpal
Nov 8 at 10:09
Are you aware of the combination/selection formula?
– Prakhar Nagpal
Nov 8 at 10:17
add a comment |
3
Please try writing the question out using mathjax math.meta.stackexchange.com/questions/5020/…. Also only enter one question at a time.
– Prakhar Nagpal
Nov 8 at 10:09
Are you aware of the combination/selection formula?
– Prakhar Nagpal
Nov 8 at 10:17
3
3
Please try writing the question out using mathjax math.meta.stackexchange.com/questions/5020/…. Also only enter one question at a time.
– Prakhar Nagpal
Nov 8 at 10:09
Please try writing the question out using mathjax math.meta.stackexchange.com/questions/5020/…. Also only enter one question at a time.
– Prakhar Nagpal
Nov 8 at 10:09
Are you aware of the combination/selection formula?
– Prakhar Nagpal
Nov 8 at 10:17
Are you aware of the combination/selection formula?
– Prakhar Nagpal
Nov 8 at 10:17
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
So, I will answer the second question for you, and also I highly recommend you provide some sort of input as to whether you have learned basic problem-solving in Combinatorics.
Now, the question says that there are $4$ friends, say $A, B, C, D$ and there are $8$ distinct books to be given to them, such that the first $2, A$ and $B$ have $4$ books in total. So, first, we need to select $4$ out of the $8$ books to be given to the $2$ friends and there are $binom 8 4$ ways of doing that.
Also then we have to give the books out to the $2$ people, and since we have $2$ options for each book, we have $2^4$ ways of giving the books to the first $2$ friends.
Next, we have to give the remaining $4$ to the other $2$ and this can also be done in $2^4$ ways since we can give each book to one of them. So finally in conclusion you get,
$$binom 8 4 cdot 2^4 cdot 2^4$$
answered Nov 8 at 10:15
Prakhar Nagpal
596318
Okey, I think I understand the solution. Thank you so much. Additionally, you told me the logic behind the solution. Are there any formula to not lost in that logic, directly apply the formula and get the result for this kind of questions ?
– Ozan
Nov 8 at 10:23
You're welcome. For giving out any $m$ distinct objects to $n$ people the formula will always be $n^m$, however, I prefer not to remember formulae, rather you will appreciate and understand with time that each good question in Combinatorics will require you to solve with logic and no one formula will be readily applicable. Also if you like the answer you can upvote it... xD
– Prakhar Nagpal
Nov 8 at 10:29
Yes. Let’s assume you have $n$ choose $r$ options. For permutations with repetition, you’ll have $n^r$ permutations. For permutations without repetition, you’ll have $P_{(n, r)} = frac{n!}{(n-r)!}$ permutations. For combinations without repetition, you’ll have ${n choose r} = frac{n!}{r!(n-r)!}$ combinations. However, as mentioned, knowing the logic helps much more than remembering the formulas. (You reach the formulas through the logic anyway.)
– KM101
Nov 8 at 10:32
add a comment |
up vote
0
down vote
Answer 1)
Total possible outcome=(total no.of options)^no. of times taken. So, it is $3^7$.
No. of combination$=binom nr$. So favourable outcomes for one student $=binom72$. So for 2 students, it is $2^{7-2}$ chances. So the result is $frac{binom72 2^5}{3^7}$.
edited Nov 18 at 3:13
Parcly Taxel
41k137198
answered Nov 8 at 10:22
Mridul Basu
11
Please use Mathjax math.meta.stackexchange.com/questions/5020/…
– Prakhar Nagpal
Nov 8 at 10:23
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
So, I will answer the second question for you, and also I highly recommend you provide some sort of input as to whether you have learned basic problem-solving in Combinatorics.
Now, the question says that there are $4$ friends, say $A, B, C, D$ and there are $8$ distinct books to be given to them, such that the first $2, A$ and $B$ have $4$ books in total. So, first, we need to select $4$ out of the $8$ books to be given to the $2$ friends and there are $binom 8 4$ ways of doing that.
Also then we have to give the books out to the $2$ people, and since we have $2$ options for each book, we have $2^4$ ways of giving the books to the first $2$ friends.
Next, we have to give the remaining $4$ to the other $2$ and this can also be done in $2^4$ ways since we can give each book to one of them. So finally in conclusion you get,
$$binom 8 4 cdot 2^4 cdot 2^4$$
answered Nov 8 at 10:15
Prakhar Nagpal
596318
Okey, I think I understand the solution. Thank you so much. Additionally, you told me the logic behind the solution. Are there any formula to not lost in that logic, directly apply the formula and get the result for this kind of questions ?
– Ozan
Nov 8 at 10:23
You're welcome. For giving out any $m$ distinct objects to $n$ people the formula will always be $n^m$, however, I prefer not to remember formulae, rather you will appreciate and understand with time that each good question in Combinatorics will require you to solve with logic and no one formula will be readily applicable. Also if you like the answer you can upvote it... xD
– Prakhar Nagpal
Nov 8 at 10:29
Yes. Let’s assume you have $n$ choose $r$ options. For permutations with repetition, you’ll have $n^r$ permutations. For permutations without repetition, you’ll have $P_{(n, r)} = frac{n!}{(n-r)!}$ permutations. For combinations without repetition, you’ll have ${n choose r} = frac{n!}{r!(n-r)!}$ combinations. However, as mentioned, knowing the logic helps much more than remembering the formulas. (You reach the formulas through the logic anyway.)
– KM101
Nov 8 at 10:32
add a comment |
up vote
0
down vote
accepted
So, I will answer the second question for you, and also I highly recommend you provide some sort of input as to whether you have learned basic problem-solving in Combinatorics.
Now, the question says that there are $4$ friends, say $A, B, C, D$ and there are $8$ distinct books to be given to them, such that the first $2, A$ and $B$ have $4$ books in total. So, first, we need to select $4$ out of the $8$ books to be given to the $2$ friends and there are $binom 8 4$ ways of doing that.
Also then we have to give the books out to the $2$ people, and since we have $2$ options for each book, we have $2^4$ ways of giving the books to the first $2$ friends.
Next, we have to give the remaining $4$ to the other $2$ and this can also be done in $2^4$ ways since we can give each book to one of them. So finally in conclusion you get,
$$binom 8 4 cdot 2^4 cdot 2^4$$
answered Nov 8 at 10:15
Prakhar Nagpal
596318
Okey, I think I understand the solution. Thank you so much. Additionally, you told me the logic behind the solution. Are there any formula to not lost in that logic, directly apply the formula and get the result for this kind of questions ?
– Ozan
Nov 8 at 10:23
You're welcome. For giving out any $m$ distinct objects to $n$ people the formula will always be $n^m$, however, I prefer not to remember formulae, rather you will appreciate and understand with time that each good question in Combinatorics will require you to solve with logic and no one formula will be readily applicable. Also if you like the answer you can upvote it... xD
– Prakhar Nagpal
Nov 8 at 10:29
Yes. Let’s assume you have $n$ choose $r$ options. For permutations with repetition, you’ll have $n^r$ permutations. For permutations without repetition, you’ll have $P_{(n, r)} = frac{n!}{(n-r)!}$ permutations. For combinations without repetition, you’ll have ${n choose r} = frac{n!}{r!(n-r)!}$ combinations. However, as mentioned, knowing the logic helps much more than remembering the formulas. (You reach the formulas through the logic anyway.)
– KM101
Nov 8 at 10:32
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
So, I will answer the second question for you, and also I highly recommend you provide some sort of input as to whether you have learned basic problem-solving in Combinatorics.
Now, the question says that there are $4$ friends, say $A, B, C, D$ and there are $8$ distinct books to be given to them, such that the first $2, A$ and $B$ have $4$ books in total. So, first, we need to select $4$ out of the $8$ books to be given to the $2$ friends and there are $binom 8 4$ ways of doing that.
Also then we have to give the books out to the $2$ people, and since we have $2$ options for each book, we have $2^4$ ways of giving the books to the first $2$ friends.
Next, we have to give the remaining $4$ to the other $2$ and this can also be done in $2^4$ ways since we can give each book to one of them. So finally in conclusion you get,
$$binom 8 4 cdot 2^4 cdot 2^4$$
answered Nov 8 at 10:15
Prakhar Nagpal
596318
So, I will answer the second question for you, and also I highly recommend you provide some sort of input as to whether you have learned basic problem-solving in Combinatorics.
Now, the question says that there are $4$ friends, say $A, B, C, D$ and there are $8$ distinct books to be given to them, such that the first $2, A$ and $B$ have $4$ books in total. So, first, we need to select $4$ out of the $8$ books to be given to the $2$ friends and there are $binom 8 4$ ways of doing that.
Also then we have to give the books out to the $2$ people, and since we have $2$ options for each book, we have $2^4$ ways of giving the books to the first $2$ friends.
Next, we have to give the remaining $4$ to the other $2$ and this can also be done in $2^4$ ways since we can give each book to one of them. So finally in conclusion you get,
$$binom 8 4 cdot 2^4 cdot 2^4$$
answered Nov 8 at 10:15
Prakhar Nagpal
596318
answered Nov 8 at 10:15
Prakhar Nagpal
596318
answered Nov 8 at 10:15
Prakhar Nagpal
596318
answered Nov 8 at 10:15
Prakhar Nagpal
596318
596318
Okey, I think I understand the solution. Thank you so much. Additionally, you told me the logic behind the solution. Are there any formula to not lost in that logic, directly apply the formula and get the result for this kind of questions ?
– Ozan
Nov 8 at 10:23
You're welcome. For giving out any $m$ distinct objects to $n$ people the formula will always be $n^m$, however, I prefer not to remember formulae, rather you will appreciate and understand with time that each good question in Combinatorics will require you to solve with logic and no one formula will be readily applicable. Also if you like the answer you can upvote it... xD
– Prakhar Nagpal
Nov 8 at 10:29
Yes. Let’s assume you have $n$ choose $r$ options. For permutations with repetition, you’ll have $n^r$ permutations. For permutations without repetition, you’ll have $P_{(n, r)} = frac{n!}{(n-r)!}$ permutations. For combinations without repetition, you’ll have ${n choose r} = frac{n!}{r!(n-r)!}$ combinations. However, as mentioned, knowing the logic helps much more than remembering the formulas. (You reach the formulas through the logic anyway.)
– KM101
Nov 8 at 10:32
add a comment |
Okey, I think I understand the solution. Thank you so much. Additionally, you told me the logic behind the solution. Are there any formula to not lost in that logic, directly apply the formula and get the result for this kind of questions ?
– Ozan
Nov 8 at 10:23
You're welcome. For giving out any $m$ distinct objects to $n$ people the formula will always be $n^m$, however, I prefer not to remember formulae, rather you will appreciate and understand with time that each good question in Combinatorics will require you to solve with logic and no one formula will be readily applicable. Also if you like the answer you can upvote it... xD
– Prakhar Nagpal
Nov 8 at 10:29
Yes. Let’s assume you have $n$ choose $r$ options. For permutations with repetition, you’ll have $n^r$ permutations. For permutations without repetition, you’ll have $P_{(n, r)} = frac{n!}{(n-r)!}$ permutations. For combinations without repetition, you’ll have ${n choose r} = frac{n!}{r!(n-r)!}$ combinations. However, as mentioned, knowing the logic helps much more than remembering the formulas. (You reach the formulas through the logic anyway.)
– KM101
Nov 8 at 10:32
Okey, I think I understand the solution. Thank you so much. Additionally, you told me the logic behind the solution. Are there any formula to not lost in that logic, directly apply the formula and get the result for this kind of questions ?
– Ozan
Nov 8 at 10:23
Okey, I think I understand the solution. Thank you so much. Additionally, you told me the logic behind the solution. Are there any formula to not lost in that logic, directly apply the formula and get the result for this kind of questions ?
– Ozan
Nov 8 at 10:23
You're welcome. For giving out any $m$ distinct objects to $n$ people the formula will always be $n^m$, however, I prefer not to remember formulae, rather you will appreciate and understand with time that each good question in Combinatorics will require you to solve with logic and no one formula will be readily applicable. Also if you like the answer you can upvote it... xD
– Prakhar Nagpal
Nov 8 at 10:29
You're welcome. For giving out any $m$ distinct objects to $n$ people the formula will always be $n^m$, however, I prefer not to remember formulae, rather you will appreciate and understand with time that each good question in Combinatorics will require you to solve with logic and no one formula will be readily applicable. Also if you like the answer you can upvote it... xD
– Prakhar Nagpal
Nov 8 at 10:29
Yes. Let’s assume you have $n$ choose $r$ options. For permutations with repetition, you’ll have $n^r$ permutations. For permutations without repetition, you’ll have $P_{(n, r)} = frac{n!}{(n-r)!}$ permutations. For combinations without repetition, you’ll have ${n choose r} = frac{n!}{r!(n-r)!}$ combinations. However, as mentioned, knowing the logic helps much more than remembering the formulas. (You reach the formulas through the logic anyway.)
– KM101
Nov 8 at 10:32
Yes. Let’s assume you have $n$ choose $r$ options. For permutations with repetition, you’ll have $n^r$ permutations. For permutations without repetition, you’ll have $P_{(n, r)} = frac{n!}{(n-r)!}$ permutations. For combinations without repetition, you’ll have ${n choose r} = frac{n!}{r!(n-r)!}$ combinations. However, as mentioned, knowing the logic helps much more than remembering the formulas. (You reach the formulas through the logic anyway.)
– KM101
Nov 8 at 10:32
add a comment |
up vote
0
down vote
Answer 1)
Total possible outcome=(total no.of options)^no. of times taken. So, it is $3^7$.
No. of combination$=binom nr$. So favourable outcomes for one student $=binom72$. So for 2 students, it is $2^{7-2}$ chances. So the result is $frac{binom72 2^5}{3^7}$.
edited Nov 18 at 3:13
Parcly Taxel
41k137198
answered Nov 8 at 10:22
Mridul Basu
11
Please use Mathjax math.meta.stackexchange.com/questions/5020/…
– Prakhar Nagpal
Nov 8 at 10:23
add a comment |
up vote
0
down vote
Answer 1)
Total possible outcome=(total no.of options)^no. of times taken. So, it is $3^7$.
No. of combination$=binom nr$. So favourable outcomes for one student $=binom72$. So for 2 students, it is $2^{7-2}$ chances. So the result is $frac{binom72 2^5}{3^7}$.
edited Nov 18 at 3:13
Parcly Taxel
41k137198
answered Nov 8 at 10:22
Mridul Basu
11
Please use Mathjax math.meta.stackexchange.com/questions/5020/…
– Prakhar Nagpal
Nov 8 at 10:23
add a comment |
up vote
0
down vote
up vote
0
down vote
Answer 1)
Total possible outcome=(total no.of options)^no. of times taken. So, it is $3^7$.
No. of combination$=binom nr$. So favourable outcomes for one student $=binom72$. So for 2 students, it is $2^{7-2}$ chances. So the result is $frac{binom72 2^5}{3^7}$.
edited Nov 18 at 3:13
Parcly Taxel
41k137198
answered Nov 8 at 10:22
Mridul Basu
11
Answer 1)
Total possible outcome=(total no.of options)^no. of times taken. So, it is $3^7$.
No. of combination$=binom nr$. So favourable outcomes for one student $=binom72$. So for 2 students, it is $2^{7-2}$ chances. So the result is $frac{binom72 2^5}{3^7}$.
edited Nov 18 at 3:13
Parcly Taxel
41k137198
answered Nov 8 at 10:22
Mridul Basu
11
edited Nov 18 at 3:13
Parcly Taxel
41k137198
edited Nov 18 at 3:13
Parcly Taxel
41k137198
edited Nov 18 at 3:13
Parcly Taxel
41k137198
41k137198
answered Nov 8 at 10:22
Mridul Basu
11
answered Nov 8 at 10:22
Mridul Basu
11
answered Nov 8 at 10:22
Mridul Basu
11
11
Please use Mathjax math.meta.stackexchange.com/questions/5020/…
– Prakhar Nagpal
Nov 8 at 10:23
add a comment |
Please use Mathjax math.meta.stackexchange.com/questions/5020/…
– Prakhar Nagpal
Nov 8 at 10:23
Please use Mathjax math.meta.stackexchange.com/questions/5020/…
– Prakhar Nagpal
Nov 8 at 10:23
Please use Mathjax math.meta.stackexchange.com/questions/5020/…
– Prakhar Nagpal
Nov 8 at 10:23
add a comment |
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Please try writing the question out using mathjax math.meta.stackexchange.com/questions/5020/…. Also only enter one question at a time.
– Prakhar Nagpal
Nov 8 at 10:09
Are you aware of the combination/selection formula?
– Prakhar Nagpal
Nov 8 at 10:17