probability of having specific numbers in an array selected from another array
$begingroup$
Let's assume that A = {1,2,3,4,5,6}.
3 numbers are selected from A in a repeated manner. Therefore, there are 6x6x6=216 possibilities which are;
111,112,113,...,166,
211,212,213,...,266,
311,312,313,...,366,
411,412,413,...,466,
511,512,513,...,566,
611,612,613,...,666.
How many of them contain 2 AND 5 ?
When calculated manually, the result is 30. However, how to get 30 by calculating with permutation and combination?
For this type of questions, I want to extract a general formula with permutation and combination.
Thanks in advance.
conditional-probability
asked Dec 18 '18 at 15:06
TalhaTalha
11
$endgroup$
add a comment |
$begingroup$
Let's assume that A = {1,2,3,4,5,6}.
3 numbers are selected from A in a repeated manner. Therefore, there are 6x6x6=216 possibilities which are;
111,112,113,...,166,
211,212,213,...,266,
311,312,313,...,366,
411,412,413,...,466,
511,512,513,...,566,
611,612,613,...,666.
How many of them contain 2 AND 5 ?
When calculated manually, the result is 30. However, how to get 30 by calculating with permutation and combination?
For this type of questions, I want to extract a general formula with permutation and combination.
Thanks in advance.
conditional-probability
asked Dec 18 '18 at 15:06
TalhaTalha
11
$endgroup$
add a comment |
$begingroup$
Let's assume that A = {1,2,3,4,5,6}.
3 numbers are selected from A in a repeated manner. Therefore, there are 6x6x6=216 possibilities which are;
111,112,113,...,166,
211,212,213,...,266,
311,312,313,...,366,
411,412,413,...,466,
511,512,513,...,566,
611,612,613,...,666.
How many of them contain 2 AND 5 ?
When calculated manually, the result is 30. However, how to get 30 by calculating with permutation and combination?
For this type of questions, I want to extract a general formula with permutation and combination.
Thanks in advance.
conditional-probability
asked Dec 18 '18 at 15:06
TalhaTalha
11
$endgroup$
Let's assume that A = {1,2,3,4,5,6}.
3 numbers are selected from A in a repeated manner. Therefore, there are 6x6x6=216 possibilities which are;
111,112,113,...,166,
211,212,213,...,266,
311,312,313,...,366,
411,412,413,...,466,
511,512,513,...,566,
611,612,613,...,666.
How many of them contain 2 AND 5 ?
When calculated manually, the result is 30. However, how to get 30 by calculating with permutation and combination?
For this type of questions, I want to extract a general formula with permutation and combination.
Thanks in advance.
conditional-probability
conditional-probability
asked Dec 18 '18 at 15:06
TalhaTalha
11
asked Dec 18 '18 at 15:06
TalhaTalha
11
edited Dec 19 '18 at 10:52
Talha
asked Dec 18 '18 at 15:06
TalhaTalha
11
asked Dec 18 '18 at 15:06
TalhaTalha
11
asked Dec 18 '18 at 15:06
TalhaTalha
11
11
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that there are 6 ways to choose the placement of the roll of 2 and 5 among the three rolls, and there are 6 choices for the remaining rolls. Hence, there are a total of 36 possibilities.
However, we have overcounted the case when the third roll is either 2 or 5. Note that if the third roll is 2, there are 3 possibilities of how to arrange the three rolls, as is the case when the third roll is 5.
Hence, our answer is $36-3-3=30$.
answered Dec 18 '18 at 15:14
Don ThousandDon Thousand
4,513734
$endgroup$
$begingroup$
It looks like it is right. But, is it possible to get 30 with a general formula in which permutation and combination is used? When the numbers are large, it will be difficult to think all possibilities and overcounted ones.
$endgroup$
– Talha
Dec 19 '18 at 8:42
$begingroup$
Nope, you have to use inclusion-exclusion unfortunately.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:08
add a comment |
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1 Answer
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$begingroup$
Note that there are 6 ways to choose the placement of the roll of 2 and 5 among the three rolls, and there are 6 choices for the remaining rolls. Hence, there are a total of 36 possibilities.
However, we have overcounted the case when the third roll is either 2 or 5. Note that if the third roll is 2, there are 3 possibilities of how to arrange the three rolls, as is the case when the third roll is 5.
Hence, our answer is $36-3-3=30$.
answered Dec 18 '18 at 15:14
Don ThousandDon Thousand
4,513734
$endgroup$
$begingroup$
It looks like it is right. But, is it possible to get 30 with a general formula in which permutation and combination is used? When the numbers are large, it will be difficult to think all possibilities and overcounted ones.
$endgroup$
– Talha
Dec 19 '18 at 8:42
$begingroup$
Nope, you have to use inclusion-exclusion unfortunately.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:08
add a comment |
$begingroup$
Note that there are 6 ways to choose the placement of the roll of 2 and 5 among the three rolls, and there are 6 choices for the remaining rolls. Hence, there are a total of 36 possibilities.
However, we have overcounted the case when the third roll is either 2 or 5. Note that if the third roll is 2, there are 3 possibilities of how to arrange the three rolls, as is the case when the third roll is 5.
Hence, our answer is $36-3-3=30$.
answered Dec 18 '18 at 15:14
Don ThousandDon Thousand
4,513734
$endgroup$
$begingroup$
It looks like it is right. But, is it possible to get 30 with a general formula in which permutation and combination is used? When the numbers are large, it will be difficult to think all possibilities and overcounted ones.
$endgroup$
– Talha
Dec 19 '18 at 8:42
$begingroup$
Nope, you have to use inclusion-exclusion unfortunately.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:08
add a comment |
$begingroup$
Note that there are 6 ways to choose the placement of the roll of 2 and 5 among the three rolls, and there are 6 choices for the remaining rolls. Hence, there are a total of 36 possibilities.
However, we have overcounted the case when the third roll is either 2 or 5. Note that if the third roll is 2, there are 3 possibilities of how to arrange the three rolls, as is the case when the third roll is 5.
Hence, our answer is $36-3-3=30$.
answered Dec 18 '18 at 15:14
Don ThousandDon Thousand
4,513734
$endgroup$
Note that there are 6 ways to choose the placement of the roll of 2 and 5 among the three rolls, and there are 6 choices for the remaining rolls. Hence, there are a total of 36 possibilities.
However, we have overcounted the case when the third roll is either 2 or 5. Note that if the third roll is 2, there are 3 possibilities of how to arrange the three rolls, as is the case when the third roll is 5.
Hence, our answer is $36-3-3=30$.
answered Dec 18 '18 at 15:14
Don ThousandDon Thousand
4,513734
answered Dec 18 '18 at 15:14
Don ThousandDon Thousand
4,513734
answered Dec 18 '18 at 15:14
Don ThousandDon Thousand
4,513734
answered Dec 18 '18 at 15:14
Don ThousandDon Thousand
4,513734
4,513734
$begingroup$
It looks like it is right. But, is it possible to get 30 with a general formula in which permutation and combination is used? When the numbers are large, it will be difficult to think all possibilities and overcounted ones.
$endgroup$
– Talha
Dec 19 '18 at 8:42
$begingroup$
Nope, you have to use inclusion-exclusion unfortunately.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:08
add a comment |
$begingroup$
It looks like it is right. But, is it possible to get 30 with a general formula in which permutation and combination is used? When the numbers are large, it will be difficult to think all possibilities and overcounted ones.
$endgroup$
– Talha
Dec 19 '18 at 8:42
$begingroup$
Nope, you have to use inclusion-exclusion unfortunately.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:08
$begingroup$
It looks like it is right. But, is it possible to get 30 with a general formula in which permutation and combination is used? When the numbers are large, it will be difficult to think all possibilities and overcounted ones.
$endgroup$
– Talha
Dec 19 '18 at 8:42
$begingroup$
It looks like it is right. But, is it possible to get 30 with a general formula in which permutation and combination is used? When the numbers are large, it will be difficult to think all possibilities and overcounted ones.
$endgroup$
– Talha
Dec 19 '18 at 8:42
$begingroup$
Nope, you have to use inclusion-exclusion unfortunately.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:08
$begingroup$
Nope, you have to use inclusion-exclusion unfortunately.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:08
add a comment |
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