Mixing
Explain the difference between these two questions, because their answers are different.
$begingroup$
Q1- What is the total number of ways of selecting at least one object from $2$ sets of $12$ different objects, each set contains $6$ objects?
The answer is $4095$.
Q2- What is the total number of ways of selecting at least one object from $2$ different sets, each set containing $6$ identical objects?
The answer is $48$.
combinatorics combinations
edited Feb 12 at 17:08
Dr. Mathva
3,190529
asked Jan 30 at 9:48
Aayush methiAayush methi
63
$endgroup$
add a comment |
$begingroup$
Q1- What is the total number of ways of selecting at least one object from $2$ sets of $12$ different objects, each set contains $6$ objects?
The answer is $4095$.
Q2- What is the total number of ways of selecting at least one object from $2$ different sets, each set containing $6$ identical objects?
The answer is $48$.
combinatorics combinations
edited Feb 12 at 17:08
Dr. Mathva
3,190529
asked Jan 30 at 9:48
Aayush methiAayush methi
63
$endgroup$
1
$begingroup$
The questions are different. Perhaps you could explain why you thought the answers would be the same.
$endgroup$
– N. F. Taussig
Jan 30 at 9:56
$begingroup$
No I didn't thought the answer would be same. Actually I couldn't find out difference in solving them.
$endgroup$
– Aayush methi
Jan 30 at 11:03
add a comment |
$begingroup$
Q1- What is the total number of ways of selecting at least one object from $2$ sets of $12$ different objects, each set contains $6$ objects?
The answer is $4095$.
Q2- What is the total number of ways of selecting at least one object from $2$ different sets, each set containing $6$ identical objects?
The answer is $48$.
combinatorics combinations
edited Feb 12 at 17:08
Dr. Mathva
3,190529
asked Jan 30 at 9:48
Aayush methiAayush methi
63
$endgroup$
Q1- What is the total number of ways of selecting at least one object from $2$ sets of $12$ different objects, each set contains $6$ objects?
The answer is $4095$.
Q2- What is the total number of ways of selecting at least one object from $2$ different sets, each set containing $6$ identical objects?
The answer is $48$.
combinatorics combinations
combinatorics combinations
edited Feb 12 at 17:08
Dr. Mathva
3,190529
asked Jan 30 at 9:48
Aayush methiAayush methi
63
edited Feb 12 at 17:08
Dr. Mathva
3,190529
asked Jan 30 at 9:48
Aayush methiAayush methi
63
edited Feb 12 at 17:08
Dr. Mathva
3,190529
edited Feb 12 at 17:08
Dr. Mathva
3,190529
edited Feb 12 at 17:08
Dr. Mathva
3,190529
3,190529
asked Jan 30 at 9:48
Aayush methiAayush methi
63
asked Jan 30 at 9:48
Aayush methiAayush methi
63
asked Jan 30 at 9:48
Aayush methiAayush methi
63
63
1
$begingroup$
The questions are different. Perhaps you could explain why you thought the answers would be the same.
$endgroup$
– N. F. Taussig
Jan 30 at 9:56
$begingroup$
No I didn't thought the answer would be same. Actually I couldn't find out difference in solving them.
$endgroup$
– Aayush methi
Jan 30 at 11:03
add a comment |
1
$begingroup$
The questions are different. Perhaps you could explain why you thought the answers would be the same.
$endgroup$
– N. F. Taussig
Jan 30 at 9:56
$begingroup$
No I didn't thought the answer would be same. Actually I couldn't find out difference in solving them.
$endgroup$
– Aayush methi
Jan 30 at 11:03
1
1
$begingroup$
The questions are different. Perhaps you could explain why you thought the answers would be the same.
$endgroup$
– N. F. Taussig
Jan 30 at 9:56
$begingroup$
The questions are different. Perhaps you could explain why you thought the answers would be the same.
$endgroup$
– N. F. Taussig
Jan 30 at 9:56
$begingroup$
No I didn't thought the answer would be same. Actually I couldn't find out difference in solving them.
$endgroup$
– Aayush methi
Jan 30 at 11:03
$begingroup$
No I didn't thought the answer would be same. Actually I couldn't find out difference in solving them.
$endgroup$
– Aayush methi
Jan 30 at 11:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The difference is that in the first question the objects in a set are different, so it matters not only how many you take, but which ones. In the second question the objects in each set are the same, so all that matters is how many you take.
answered Jan 30 at 10:04
Especially LimeEspecially Lime
22.7k23059
$endgroup$
$begingroup$
Oh I got it now. Can you please explain how we got 48 as the answer of question 2.
$endgroup$
– Aayush methi
Jan 30 at 11:04
1
$begingroup$
@Aayushmethi You have $7$ options for each set ($0$ to $6$ objects selected) with $1$ invalid selection $(0+0)$ This gives $7times7-1=48$
$endgroup$
– Daniel Mathias
Jan 30 at 13:38
$begingroup$
OK. Thanks a lot.
$endgroup$
– Aayush methi
Jan 31 at 5:43
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The difference is that in the first question the objects in a set are different, so it matters not only how many you take, but which ones. In the second question the objects in each set are the same, so all that matters is how many you take.
answered Jan 30 at 10:04
Especially LimeEspecially Lime
22.7k23059
$endgroup$
$begingroup$
Oh I got it now. Can you please explain how we got 48 as the answer of question 2.
$endgroup$
– Aayush methi
Jan 30 at 11:04
1
$begingroup$
@Aayushmethi You have $7$ options for each set ($0$ to $6$ objects selected) with $1$ invalid selection $(0+0)$ This gives $7times7-1=48$
$endgroup$
– Daniel Mathias
Jan 30 at 13:38
$begingroup$
OK. Thanks a lot.
$endgroup$
– Aayush methi
Jan 31 at 5:43
add a comment |
$begingroup$
The difference is that in the first question the objects in a set are different, so it matters not only how many you take, but which ones. In the second question the objects in each set are the same, so all that matters is how many you take.
answered Jan 30 at 10:04
Especially LimeEspecially Lime
22.7k23059
$endgroup$
$begingroup$
Oh I got it now. Can you please explain how we got 48 as the answer of question 2.
$endgroup$
– Aayush methi
Jan 30 at 11:04
1
$begingroup$
@Aayushmethi You have $7$ options for each set ($0$ to $6$ objects selected) with $1$ invalid selection $(0+0)$ This gives $7times7-1=48$
$endgroup$
– Daniel Mathias
Jan 30 at 13:38
$begingroup$
OK. Thanks a lot.
$endgroup$
– Aayush methi
Jan 31 at 5:43
add a comment |
$begingroup$
The difference is that in the first question the objects in a set are different, so it matters not only how many you take, but which ones. In the second question the objects in each set are the same, so all that matters is how many you take.
answered Jan 30 at 10:04
Especially LimeEspecially Lime
22.7k23059
$endgroup$
The difference is that in the first question the objects in a set are different, so it matters not only how many you take, but which ones. In the second question the objects in each set are the same, so all that matters is how many you take.
answered Jan 30 at 10:04
Especially LimeEspecially Lime
22.7k23059
answered Jan 30 at 10:04
Especially LimeEspecially Lime
22.7k23059
answered Jan 30 at 10:04
Especially LimeEspecially Lime
22.7k23059
answered Jan 30 at 10:04
Especially LimeEspecially Lime
22.7k23059
22.7k23059
$begingroup$
Oh I got it now. Can you please explain how we got 48 as the answer of question 2.
$endgroup$
– Aayush methi
Jan 30 at 11:04
1
$begingroup$
@Aayushmethi You have $7$ options for each set ($0$ to $6$ objects selected) with $1$ invalid selection $(0+0)$ This gives $7times7-1=48$
$endgroup$
– Daniel Mathias
Jan 30 at 13:38
$begingroup$
OK. Thanks a lot.
$endgroup$
– Aayush methi
Jan 31 at 5:43
add a comment |
$begingroup$
Oh I got it now. Can you please explain how we got 48 as the answer of question 2.
$endgroup$
– Aayush methi
Jan 30 at 11:04
1
$begingroup$
@Aayushmethi You have $7$ options for each set ($0$ to $6$ objects selected) with $1$ invalid selection $(0+0)$ This gives $7times7-1=48$
$endgroup$
– Daniel Mathias
Jan 30 at 13:38
$begingroup$
OK. Thanks a lot.
$endgroup$
– Aayush methi
Jan 31 at 5:43
$begingroup$
Oh I got it now. Can you please explain how we got 48 as the answer of question 2.
$endgroup$
– Aayush methi
Jan 30 at 11:04
$begingroup$
Oh I got it now. Can you please explain how we got 48 as the answer of question 2.
$endgroup$
– Aayush methi
Jan 30 at 11:04
1
1
$begingroup$
@Aayushmethi You have $7$ options for each set ($0$ to $6$ objects selected) with $1$ invalid selection $(0+0)$ This gives $7times7-1=48$
$endgroup$
– Daniel Mathias
Jan 30 at 13:38
$begingroup$
@Aayushmethi You have $7$ options for each set ($0$ to $6$ objects selected) with $1$ invalid selection $(0+0)$ This gives $7times7-1=48$
$endgroup$
– Daniel Mathias
Jan 30 at 13:38
$begingroup$
OK. Thanks a lot.
$endgroup$
– Aayush methi
Jan 31 at 5:43
$begingroup$
OK. Thanks a lot.
$endgroup$
– Aayush methi
Jan 31 at 5:43
add a comment |
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1
$begingroup$
The questions are different. Perhaps you could explain why you thought the answers would be the same.
$endgroup$
– N. F. Taussig
Jan 30 at 9:56
$begingroup$
No I didn't thought the answer would be same. Actually I couldn't find out difference in solving them.
$endgroup$
– Aayush methi
Jan 30 at 11:03