Need help with the combinations
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Allow me to ask (may be a silly) question, I got my school exam tommorow and I'm stuck in a question. (Orignal question is different)
There is a bag with 9 balls including 3 blue , 2 red , 4 black balls. Balls of same color are non identical, we have to select 4 balls such that there must be atleast a ball of each color.
I deduce answer from original question to be 144.
And I did get 144
It went like
Selecting one ball of each color amd then selecting 1 ball from left 6 balls.
$$^4 C_1 × ^3 C_1× ^2 C_1× ^6 C_1$$ which gives 144.
If the above math doesn't look the way I expected, then it is
4C1 × 3C1× 2C1 ×6C1
But, what confuses me is,
I made 3 cases.
First selected 2 balls from blue , one each from rest two, then did the same 3 times with other colors. Which added upto 72.
Answer should be 144.
I don't know where I am lacking, is it a conceptual error?
combinatorics combinations
edited Nov 15 at 16:35
N. F. Taussig
42.4k93254
asked Nov 15 at 13:58
Chirag Mehta
343
|
show 1 more comment
up vote
0
down vote
favorite
Allow me to ask (may be a silly) question, I got my school exam tommorow and I'm stuck in a question. (Orignal question is different)
There is a bag with 9 balls including 3 blue , 2 red , 4 black balls. Balls of same color are non identical, we have to select 4 balls such that there must be atleast a ball of each color.
I deduce answer from original question to be 144.
And I did get 144
It went like
Selecting one ball of each color amd then selecting 1 ball from left 6 balls.
$$^4 C_1 × ^3 C_1× ^2 C_1× ^6 C_1$$ which gives 144.
If the above math doesn't look the way I expected, then it is
4C1 × 3C1× 2C1 ×6C1
But, what confuses me is,
I made 3 cases.
First selected 2 balls from blue , one each from rest two, then did the same 3 times with other colors. Which added upto 72.
Answer should be 144.
I don't know where I am lacking, is it a conceptual error?
combinatorics combinations
edited Nov 15 at 16:35
N. F. Taussig
42.4k93254
asked Nov 15 at 13:58
Chirag Mehta
343
One downvote immediately, question has to be silly
– Chirag Mehta
Nov 15 at 14:03
1
I don't think it is silly. Let's just look at each case. Two blues: $3times 2times 4=24$. Two reds: $3times 4=12$. Two blacks: $3times 2times 6=36$ The total is $72$. The other answer, $144$, doesn't even makes sense...if we ignore the colors completely, $binom 94=126$ so the answer must be considerably smaller than that.
– lulu
Nov 15 at 14:18
2
The method you used to get $144$ double counts. The choices: $(B_1,R_1,b_1,B_2)$ and $(B_2,R_1,b_1,B_1)$ are the same selection...so you just have to divide by $2$.
– lulu
Nov 15 at 14:21
1
Good instincts by the way, to look for alternate methods. These problems can be very unintuitive, so it's a great idea to have many ways of looking at them.
– lulu
Nov 15 at 14:25
1
As yet another way to approach the problem, try Inclusion-Exclusion. There are $126$ choices if we ignore color...how many of those are missing blue? red? black? How many are missing two colors?
– lulu
Nov 15 at 14:27
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Allow me to ask (may be a silly) question, I got my school exam tommorow and I'm stuck in a question. (Orignal question is different)
There is a bag with 9 balls including 3 blue , 2 red , 4 black balls. Balls of same color are non identical, we have to select 4 balls such that there must be atleast a ball of each color.
I deduce answer from original question to be 144.
And I did get 144
It went like
Selecting one ball of each color amd then selecting 1 ball from left 6 balls.
$$^4 C_1 × ^3 C_1× ^2 C_1× ^6 C_1$$ which gives 144.
If the above math doesn't look the way I expected, then it is
4C1 × 3C1× 2C1 ×6C1
But, what confuses me is,
I made 3 cases.
First selected 2 balls from blue , one each from rest two, then did the same 3 times with other colors. Which added upto 72.
Answer should be 144.
I don't know where I am lacking, is it a conceptual error?
combinatorics combinations
edited Nov 15 at 16:35
N. F. Taussig
42.4k93254
asked Nov 15 at 13:58
Chirag Mehta
343
Allow me to ask (may be a silly) question, I got my school exam tommorow and I'm stuck in a question. (Orignal question is different)
There is a bag with 9 balls including 3 blue , 2 red , 4 black balls. Balls of same color are non identical, we have to select 4 balls such that there must be atleast a ball of each color.
I deduce answer from original question to be 144.
And I did get 144
It went like
Selecting one ball of each color amd then selecting 1 ball from left 6 balls.
$$^4 C_1 × ^3 C_1× ^2 C_1× ^6 C_1$$ which gives 144.
If the above math doesn't look the way I expected, then it is
4C1 × 3C1× 2C1 ×6C1
But, what confuses me is,
I made 3 cases.
First selected 2 balls from blue , one each from rest two, then did the same 3 times with other colors. Which added upto 72.
Answer should be 144.
I don't know where I am lacking, is it a conceptual error?
combinatorics combinations
combinatorics combinations
edited Nov 15 at 16:35
N. F. Taussig
42.4k93254
asked Nov 15 at 13:58
Chirag Mehta
343
edited Nov 15 at 16:35
N. F. Taussig
42.4k93254
asked Nov 15 at 13:58
Chirag Mehta
343
edited Nov 15 at 16:35
N. F. Taussig
42.4k93254
edited Nov 15 at 16:35
N. F. Taussig
42.4k93254
edited Nov 15 at 16:35
N. F. Taussig
42.4k93254
42.4k93254
asked Nov 15 at 13:58
Chirag Mehta
343
asked Nov 15 at 13:58
Chirag Mehta
343
asked Nov 15 at 13:58
Chirag Mehta
343
343
One downvote immediately, question has to be silly
– Chirag Mehta
Nov 15 at 14:03
1
I don't think it is silly. Let's just look at each case. Two blues: $3times 2times 4=24$. Two reds: $3times 4=12$. Two blacks: $3times 2times 6=36$ The total is $72$. The other answer, $144$, doesn't even makes sense...if we ignore the colors completely, $binom 94=126$ so the answer must be considerably smaller than that.
– lulu
Nov 15 at 14:18
2
The method you used to get $144$ double counts. The choices: $(B_1,R_1,b_1,B_2)$ and $(B_2,R_1,b_1,B_1)$ are the same selection...so you just have to divide by $2$.
– lulu
Nov 15 at 14:21
1
Good instincts by the way, to look for alternate methods. These problems can be very unintuitive, so it's a great idea to have many ways of looking at them.
– lulu
Nov 15 at 14:25
1
As yet another way to approach the problem, try Inclusion-Exclusion. There are $126$ choices if we ignore color...how many of those are missing blue? red? black? How many are missing two colors?
– lulu
Nov 15 at 14:27
|
show 1 more comment
One downvote immediately, question has to be silly
– Chirag Mehta
Nov 15 at 14:03
1
I don't think it is silly. Let's just look at each case. Two blues: $3times 2times 4=24$. Two reds: $3times 4=12$. Two blacks: $3times 2times 6=36$ The total is $72$. The other answer, $144$, doesn't even makes sense...if we ignore the colors completely, $binom 94=126$ so the answer must be considerably smaller than that.
– lulu
Nov 15 at 14:18
2
The method you used to get $144$ double counts. The choices: $(B_1,R_1,b_1,B_2)$ and $(B_2,R_1,b_1,B_1)$ are the same selection...so you just have to divide by $2$.
– lulu
Nov 15 at 14:21
1
Good instincts by the way, to look for alternate methods. These problems can be very unintuitive, so it's a great idea to have many ways of looking at them.
– lulu
Nov 15 at 14:25
1
As yet another way to approach the problem, try Inclusion-Exclusion. There are $126$ choices if we ignore color...how many of those are missing blue? red? black? How many are missing two colors?
– lulu
Nov 15 at 14:27
One downvote immediately, question has to be silly
– Chirag Mehta
Nov 15 at 14:03
One downvote immediately, question has to be silly
– Chirag Mehta
Nov 15 at 14:03
1
1
I don't think it is silly. Let's just look at each case. Two blues: $3times 2times 4=24$. Two reds: $3times 4=12$. Two blacks: $3times 2times 6=36$ The total is $72$. The other answer, $144$, doesn't even makes sense...if we ignore the colors completely, $binom 94=126$ so the answer must be considerably smaller than that.
– lulu
Nov 15 at 14:18
I don't think it is silly. Let's just look at each case. Two blues: $3times 2times 4=24$. Two reds: $3times 4=12$. Two blacks: $3times 2times 6=36$ The total is $72$. The other answer, $144$, doesn't even makes sense...if we ignore the colors completely, $binom 94=126$ so the answer must be considerably smaller than that.
– lulu
Nov 15 at 14:18
2
2
The method you used to get $144$ double counts. The choices: $(B_1,R_1,b_1,B_2)$ and $(B_2,R_1,b_1,B_1)$ are the same selection...so you just have to divide by $2$.
– lulu
Nov 15 at 14:21
The method you used to get $144$ double counts. The choices: $(B_1,R_1,b_1,B_2)$ and $(B_2,R_1,b_1,B_1)$ are the same selection...so you just have to divide by $2$.
– lulu
Nov 15 at 14:21
1
1
Good instincts by the way, to look for alternate methods. These problems can be very unintuitive, so it's a great idea to have many ways of looking at them.
– lulu
Nov 15 at 14:25
Good instincts by the way, to look for alternate methods. These problems can be very unintuitive, so it's a great idea to have many ways of looking at them.
– lulu
Nov 15 at 14:25
1
1
As yet another way to approach the problem, try Inclusion-Exclusion. There are $126$ choices if we ignore color...how many of those are missing blue? red? black? How many are missing two colors?
– lulu
Nov 15 at 14:27
As yet another way to approach the problem, try Inclusion-Exclusion. There are $126$ choices if we ignore color...how many of those are missing blue? red? black? How many are missing two colors?
– lulu
Nov 15 at 14:27
|
show 1 more comment
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One downvote immediately, question has to be silly
– Chirag Mehta
Nov 15 at 14:03
1
I don't think it is silly. Let's just look at each case. Two blues: $3times 2times 4=24$. Two reds: $3times 4=12$. Two blacks: $3times 2times 6=36$ The total is $72$. The other answer, $144$, doesn't even makes sense...if we ignore the colors completely, $binom 94=126$ so the answer must be considerably smaller than that.
– lulu
Nov 15 at 14:18
2
The method you used to get $144$ double counts. The choices: $(B_1,R_1,b_1,B_2)$ and $(B_2,R_1,b_1,B_1)$ are the same selection...so you just have to divide by $2$.
– lulu
Nov 15 at 14:21
1
Good instincts by the way, to look for alternate methods. These problems can be very unintuitive, so it's a great idea to have many ways of looking at them.
– lulu
Nov 15 at 14:25
1
As yet another way to approach the problem, try Inclusion-Exclusion. There are $126$ choices if we ignore color...how many of those are missing blue? red? black? How many are missing two colors?
– lulu
Nov 15 at 14:27