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Casework and another argument yielding contradictory results
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I am trying to solve this counting problem which asks to find the number of ways that you can select 6 balls out of 5 distinct red balls and 6 distinct white balls, with atleast 2 balls of each color.
My book claims that it is necessary to do this by casework and gives an answer 425 with the following consideration:
$$textrmNumber of ways= binom52binom64+binom53binom63+binom54binom62=425 $$
And I understand that well enough.
I came up with the following way.
There are $binom52$ ways to choose the minimum 2 balls from red and $binom62$ ways to choose the minimum 2 white balls.
Finally from the remaining 7 balls we pick two balls to finish of.
$$textrmNumber of Ways=binom52binom62binom72=21cdot 15cdot 10=3150$$
I have a feeling that I'm grossly overcounting and that there are arrangements where I say, select 2 red balls before and 2 after which is the same as another in which I choose the latter 2 before.
Is there a way to salvage this and/or another approach that is more elegant than casework?
combinatorics
asked Jul 27 at 13:56
Damien Ashwood
492
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up vote
2
down vote
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I am trying to solve this counting problem which asks to find the number of ways that you can select 6 balls out of 5 distinct red balls and 6 distinct white balls, with atleast 2 balls of each color.
My book claims that it is necessary to do this by casework and gives an answer 425 with the following consideration:
$$textrmNumber of ways= binom52binom64+binom53binom63+binom54binom62=425 $$
And I understand that well enough.
I came up with the following way.
There are $binom52$ ways to choose the minimum 2 balls from red and $binom62$ ways to choose the minimum 2 white balls.
Finally from the remaining 7 balls we pick two balls to finish of.
$$textrmNumber of Ways=binom52binom62binom72=21cdot 15cdot 10=3150$$
I have a feeling that I'm grossly overcounting and that there are arrangements where I say, select 2 red balls before and 2 after which is the same as another in which I choose the latter 2 before.
Is there a way to salvage this and/or another approach that is more elegant than casework?
combinatorics
asked Jul 27 at 13:56
Damien Ashwood
492
The cases with three red and three white balls are being counted $3times3=9$ times in you solution etc.
– Lord Shark the Unknown
Jul 27 at 14:00
Casework does fine here, and I don't see any possibilities to make things more elegant here.
– drhab
Jul 27 at 14:06
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to solve this counting problem which asks to find the number of ways that you can select 6 balls out of 5 distinct red balls and 6 distinct white balls, with atleast 2 balls of each color.
My book claims that it is necessary to do this by casework and gives an answer 425 with the following consideration:
$$textrmNumber of ways= binom52binom64+binom53binom63+binom54binom62=425 $$
And I understand that well enough.
I came up with the following way.
There are $binom52$ ways to choose the minimum 2 balls from red and $binom62$ ways to choose the minimum 2 white balls.
Finally from the remaining 7 balls we pick two balls to finish of.
$$textrmNumber of Ways=binom52binom62binom72=21cdot 15cdot 10=3150$$
I have a feeling that I'm grossly overcounting and that there are arrangements where I say, select 2 red balls before and 2 after which is the same as another in which I choose the latter 2 before.
Is there a way to salvage this and/or another approach that is more elegant than casework?
combinatorics
asked Jul 27 at 13:56
Damien Ashwood
492
I am trying to solve this counting problem which asks to find the number of ways that you can select 6 balls out of 5 distinct red balls and 6 distinct white balls, with atleast 2 balls of each color.
My book claims that it is necessary to do this by casework and gives an answer 425 with the following consideration:
$$textrmNumber of ways= binom52binom64+binom53binom63+binom54binom62=425 $$
And I understand that well enough.
I came up with the following way.
There are $binom52$ ways to choose the minimum 2 balls from red and $binom62$ ways to choose the minimum 2 white balls.
Finally from the remaining 7 balls we pick two balls to finish of.
$$textrmNumber of Ways=binom52binom62binom72=21cdot 15cdot 10=3150$$
I have a feeling that I'm grossly overcounting and that there are arrangements where I say, select 2 red balls before and 2 after which is the same as another in which I choose the latter 2 before.
Is there a way to salvage this and/or another approach that is more elegant than casework?
combinatorics
asked Jul 27 at 13:56
Damien Ashwood
492
asked Jul 27 at 13:56
Damien Ashwood
492
asked Jul 27 at 13:56
Damien Ashwood
492
asked Jul 27 at 13:56
Damien Ashwood
492
492
The cases with three red and three white balls are being counted $3times3=9$ times in you solution etc.
– Lord Shark the Unknown
Jul 27 at 14:00
Casework does fine here, and I don't see any possibilities to make things more elegant here.
– drhab
Jul 27 at 14:06
add a comment |Â
The cases with three red and three white balls are being counted $3times3=9$ times in you solution etc.
– Lord Shark the Unknown
Jul 27 at 14:00
Casework does fine here, and I don't see any possibilities to make things more elegant here.
– drhab
Jul 27 at 14:06
The cases with three red and three white balls are being counted $3times3=9$ times in you solution etc.
– Lord Shark the Unknown
Jul 27 at 14:00
The cases with three red and three white balls are being counted $3times3=9$ times in you solution etc.
– Lord Shark the Unknown
Jul 27 at 14:00
Casework does fine here, and I don't see any possibilities to make things more elegant here.
– drhab
Jul 27 at 14:06
Casework does fine here, and I don't see any possibilities to make things more elegant here.
– drhab
Jul 27 at 14:06
add a comment |Â
1 Answer
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3red/3white are counted $3times 3=9$ times as in your question & the comment. In detail: number the balls $R_1$ to $R_5$ and $W_1$ to $W_6$. Each of the following 9 cases are counted differently in your calculation when they should really be counted once:
$$R_1, R_2, R_3; W_1, W_2, W_3\
R_1, R_2, R_3; W_1, W_3, W_2\
R_1, R_2, R_3; W_2, W_3, W_1\
R_1, R_3, R_2; W_1, W_2, W_3\
R_1, R_3, R_2; W_1, W_3, W_2\
R_1, R_3, R_2; W_2, W_3, W_1\
R_2, R_3, R_1; W_1, W_2, W_3\
R_2, R_3, R_1; W_1, W_3, W_2\
R_2, R_3, R_1; W_2, W_3, W_1$$
There's nothing special about $R_1,R_2,R_3,W_1,W_2,W_3$ here; same holds for all pairs of 3red/3white.
4red/2white and 2red/4white are each counted $binom42=6$ times in your count (there's $binom42$ different ways to split $R_1, R_2, R_3, R_4$ into 2 sets of 2: the set of 2 picked before & the set of 2 picked after).
I can't really see any way other than to separate these two cases, which leads to a solution probably less elegant than the casework, but it's maybe good to see how you get to the same answer.
There's $binom53 binom63 = 200$ ways to pick 3/3.
So using the correcting-for-overcounting approach we get that the total # ways to pick 3r/3w OR 4r/2w, 2r/4w is
$$200 + frac3150-200times 96 = 200 + 225 = 425.$$
Can't think of some other more elegant solution.
answered Jul 27 at 19:30
xmq
534
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
3red/3white are counted $3times 3=9$ times as in your question & the comment. In detail: number the balls $R_1$ to $R_5$ and $W_1$ to $W_6$. Each of the following 9 cases are counted differently in your calculation when they should really be counted once:
$$R_1, R_2, R_3; W_1, W_2, W_3\
R_1, R_2, R_3; W_1, W_3, W_2\
R_1, R_2, R_3; W_2, W_3, W_1\
R_1, R_3, R_2; W_1, W_2, W_3\
R_1, R_3, R_2; W_1, W_3, W_2\
R_1, R_3, R_2; W_2, W_3, W_1\
R_2, R_3, R_1; W_1, W_2, W_3\
R_2, R_3, R_1; W_1, W_3, W_2\
R_2, R_3, R_1; W_2, W_3, W_1$$
There's nothing special about $R_1,R_2,R_3,W_1,W_2,W_3$ here; same holds for all pairs of 3red/3white.
4red/2white and 2red/4white are each counted $binom42=6$ times in your count (there's $binom42$ different ways to split $R_1, R_2, R_3, R_4$ into 2 sets of 2: the set of 2 picked before & the set of 2 picked after).
I can't really see any way other than to separate these two cases, which leads to a solution probably less elegant than the casework, but it's maybe good to see how you get to the same answer.
There's $binom53 binom63 = 200$ ways to pick 3/3.
So using the correcting-for-overcounting approach we get that the total # ways to pick 3r/3w OR 4r/2w, 2r/4w is
$$200 + frac3150-200times 96 = 200 + 225 = 425.$$
Can't think of some other more elegant solution.
answered Jul 27 at 19:30
xmq
534
add a comment |Â
up vote
0
down vote
3red/3white are counted $3times 3=9$ times as in your question & the comment. In detail: number the balls $R_1$ to $R_5$ and $W_1$ to $W_6$. Each of the following 9 cases are counted differently in your calculation when they should really be counted once:
$$R_1, R_2, R_3; W_1, W_2, W_3\
R_1, R_2, R_3; W_1, W_3, W_2\
R_1, R_2, R_3; W_2, W_3, W_1\
R_1, R_3, R_2; W_1, W_2, W_3\
R_1, R_3, R_2; W_1, W_3, W_2\
R_1, R_3, R_2; W_2, W_3, W_1\
R_2, R_3, R_1; W_1, W_2, W_3\
R_2, R_3, R_1; W_1, W_3, W_2\
R_2, R_3, R_1; W_2, W_3, W_1$$
There's nothing special about $R_1,R_2,R_3,W_1,W_2,W_3$ here; same holds for all pairs of 3red/3white.
4red/2white and 2red/4white are each counted $binom42=6$ times in your count (there's $binom42$ different ways to split $R_1, R_2, R_3, R_4$ into 2 sets of 2: the set of 2 picked before & the set of 2 picked after).
I can't really see any way other than to separate these two cases, which leads to a solution probably less elegant than the casework, but it's maybe good to see how you get to the same answer.
There's $binom53 binom63 = 200$ ways to pick 3/3.
So using the correcting-for-overcounting approach we get that the total # ways to pick 3r/3w OR 4r/2w, 2r/4w is
$$200 + frac3150-200times 96 = 200 + 225 = 425.$$
Can't think of some other more elegant solution.
answered Jul 27 at 19:30
xmq
534
add a comment |Â
up vote
0
down vote
up vote
0
down vote
3red/3white are counted $3times 3=9$ times as in your question & the comment. In detail: number the balls $R_1$ to $R_5$ and $W_1$ to $W_6$. Each of the following 9 cases are counted differently in your calculation when they should really be counted once:
$$R_1, R_2, R_3; W_1, W_2, W_3\
R_1, R_2, R_3; W_1, W_3, W_2\
R_1, R_2, R_3; W_2, W_3, W_1\
R_1, R_3, R_2; W_1, W_2, W_3\
R_1, R_3, R_2; W_1, W_3, W_2\
R_1, R_3, R_2; W_2, W_3, W_1\
R_2, R_3, R_1; W_1, W_2, W_3\
R_2, R_3, R_1; W_1, W_3, W_2\
R_2, R_3, R_1; W_2, W_3, W_1$$
There's nothing special about $R_1,R_2,R_3,W_1,W_2,W_3$ here; same holds for all pairs of 3red/3white.
4red/2white and 2red/4white are each counted $binom42=6$ times in your count (there's $binom42$ different ways to split $R_1, R_2, R_3, R_4$ into 2 sets of 2: the set of 2 picked before & the set of 2 picked after).
I can't really see any way other than to separate these two cases, which leads to a solution probably less elegant than the casework, but it's maybe good to see how you get to the same answer.
There's $binom53 binom63 = 200$ ways to pick 3/3.
So using the correcting-for-overcounting approach we get that the total # ways to pick 3r/3w OR 4r/2w, 2r/4w is
$$200 + frac3150-200times 96 = 200 + 225 = 425.$$
Can't think of some other more elegant solution.
answered Jul 27 at 19:30
xmq
534
3red/3white are counted $3times 3=9$ times as in your question & the comment. In detail: number the balls $R_1$ to $R_5$ and $W_1$ to $W_6$. Each of the following 9 cases are counted differently in your calculation when they should really be counted once:
$$R_1, R_2, R_3; W_1, W_2, W_3\
R_1, R_2, R_3; W_1, W_3, W_2\
R_1, R_2, R_3; W_2, W_3, W_1\
R_1, R_3, R_2; W_1, W_2, W_3\
R_1, R_3, R_2; W_1, W_3, W_2\
R_1, R_3, R_2; W_2, W_3, W_1\
R_2, R_3, R_1; W_1, W_2, W_3\
R_2, R_3, R_1; W_1, W_3, W_2\
R_2, R_3, R_1; W_2, W_3, W_1$$
There's nothing special about $R_1,R_2,R_3,W_1,W_2,W_3$ here; same holds for all pairs of 3red/3white.
4red/2white and 2red/4white are each counted $binom42=6$ times in your count (there's $binom42$ different ways to split $R_1, R_2, R_3, R_4$ into 2 sets of 2: the set of 2 picked before & the set of 2 picked after).
I can't really see any way other than to separate these two cases, which leads to a solution probably less elegant than the casework, but it's maybe good to see how you get to the same answer.
There's $binom53 binom63 = 200$ ways to pick 3/3.
So using the correcting-for-overcounting approach we get that the total # ways to pick 3r/3w OR 4r/2w, 2r/4w is
$$200 + frac3150-200times 96 = 200 + 225 = 425.$$
Can't think of some other more elegant solution.
answered Jul 27 at 19:30
xmq
534
edited Jul 27 at 19:35
answered Jul 27 at 19:30
xmq
534
answered Jul 27 at 19:30
xmq
534
answered Jul 27 at 19:30
xmq
534
534
add a comment |Â
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The cases with three red and three white balls are being counted $3times3=9$ times in you solution etc.
– Lord Shark the Unknown
Jul 27 at 14:00
Casework does fine here, and I don't see any possibilities to make things more elegant here.
– drhab
Jul 27 at 14:06