Mixing
What's wrong in my solution? Ways of choosing 5 items from 3 catagory with 3, 6, 14 items, while having 1 item from each catagory.
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The exact question is:
b) Sandra wishes to buy some applications (apps) for her smartphone but she only has enough
money for 5 apps in total. There are 3 train apps, 6 social network apps and 14 games apps
available. Sandra wants to have at least 1 of each type of app. Find the number of different
possible selections of 5 apps that Sandra can choose?
How I approached this question :
As we need to choose 1 from each catagory, the first three apps can be choosen in :
3 x 6 x 14 = 252 ways.
Now, for the remaining 2 apps, we can choose from the 20 apps (3 + 6 + 14 - the installed 3 apps)
So, number of combinations should be 20C2 = 190, so finally the answer I got to was 3 x 6 x 14 x 190 = 47880 which is wrong, the answer is 13839 ways. After trying another method I got 13839 but, I need to know why this method is wrong. Can anyone explain this?
combinatorics combinations
asked 37 mins ago
abnas
212
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up vote
4
down vote
favorite
The exact question is:
b) Sandra wishes to buy some applications (apps) for her smartphone but she only has enough
money for 5 apps in total. There are 3 train apps, 6 social network apps and 14 games apps
available. Sandra wants to have at least 1 of each type of app. Find the number of different
possible selections of 5 apps that Sandra can choose?
How I approached this question :
As we need to choose 1 from each catagory, the first three apps can be choosen in :
3 x 6 x 14 = 252 ways.
Now, for the remaining 2 apps, we can choose from the 20 apps (3 + 6 + 14 - the installed 3 apps)
So, number of combinations should be 20C2 = 190, so finally the answer I got to was 3 x 6 x 14 x 190 = 47880 which is wrong, the answer is 13839 ways. After trying another method I got 13839 but, I need to know why this method is wrong. Can anyone explain this?
combinatorics combinations
asked 37 mins ago
abnas
212
New contributor
abnas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The exact question is:
b) Sandra wishes to buy some applications (apps) for her smartphone but she only has enough
money for 5 apps in total. There are 3 train apps, 6 social network apps and 14 games apps
available. Sandra wants to have at least 1 of each type of app. Find the number of different
possible selections of 5 apps that Sandra can choose?
How I approached this question :
As we need to choose 1 from each catagory, the first three apps can be choosen in :
3 x 6 x 14 = 252 ways.
Now, for the remaining 2 apps, we can choose from the 20 apps (3 + 6 + 14 - the installed 3 apps)
So, number of combinations should be 20C2 = 190, so finally the answer I got to was 3 x 6 x 14 x 190 = 47880 which is wrong, the answer is 13839 ways. After trying another method I got 13839 but, I need to know why this method is wrong. Can anyone explain this?
combinatorics combinations
asked 37 mins ago
abnas
212
New contributor
abnas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The exact question is:
b) Sandra wishes to buy some applications (apps) for her smartphone but she only has enough
money for 5 apps in total. There are 3 train apps, 6 social network apps and 14 games apps
available. Sandra wants to have at least 1 of each type of app. Find the number of different
possible selections of 5 apps that Sandra can choose?
How I approached this question :
As we need to choose 1 from each catagory, the first three apps can be choosen in :
3 x 6 x 14 = 252 ways.
Now, for the remaining 2 apps, we can choose from the 20 apps (3 + 6 + 14 - the installed 3 apps)
So, number of combinations should be 20C2 = 190, so finally the answer I got to was 3 x 6 x 14 x 190 = 47880 which is wrong, the answer is 13839 ways. After trying another method I got 13839 but, I need to know why this method is wrong. Can anyone explain this?
combinatorics combinations
combinatorics combinations
asked 37 mins ago
abnas
212
New contributor
abnas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 37 mins ago
abnas
212
New contributor
abnas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 37 mins ago
abnas
212
New contributor
abnas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 37 mins ago
abnas
212
asked 37 mins ago
abnas
212
212
New contributor
abnas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
abnas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
abnas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
You are overcounting many, many choices, such as the following selections:
train app A, SNS app B, games app C, games app D, games app E
train app A, SNS app B, games app D, games app C, games app E
Your method counts these two selections as different (the italicised part is the "at least one of each app" requirement), but they are the same.
answered 21 mins ago
Parcly Taxel
35.6k136991
Ok, I see now why my answer is wrong but, can you point out where/what step in my solution this overlap occurs? Is it possible to change my solution to avoid this error?
– abnas
5 mins ago
@abnas When you start saying "the remaining two apps are selected from the unselected 20", that is where you overcount. There is no easy fix to your attempt.
– Parcly Taxel
4 mins ago
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up vote
0
down vote
As pointed out by @Parcly Taxel,
You are overcounting the apps; keep in mind that order is irrelevant here, so train1,train2 is same as train2,train1. This is where overccounting crept into your solution.
So, how to avoid this overcounting?
Take 3T,6S and 14G apps.
In total, you want 5 apps.
So make selections like: $$(1T,1S,3G),(1T,3S,1G),(3T,1S,1G),(2T,2S,1G),(2T,1S,2G),(1T,2S,2G)$$
Now, count for number of ways you have for each of the above selection, and add.
answered 4 mins ago
omega
8042818
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up vote
0
down vote
As Parcly Taxel pointed out, you overcount many times because picking the people in two different counts(using your method) doesn't account for the ordering.
My attempt(casework) is the following:
The possible group sizes per element must be either: 1, 1, 3 or 1, 2, 2.
Case 1: 1, 1, 3
There are different combinations based on the different choices of the largest group. So we compute them separately, giving
$$binom31binom61binom143 + binom31binom63binom141 + binom33binom61binom141 = 7476.$$
Case 2: 1, 2, 2
Similarly the the last one, we have
$$binom31binom62binom142+binom32binom61binom142+binom32binom62binom141 = 6363.$$
Summing the two cases gives a total of $$boxed13839.$$
answered 2 mins ago
math783625
1037
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You are overcounting many, many choices, such as the following selections:
train app A, SNS app B, games app C, games app D, games app E
train app A, SNS app B, games app D, games app C, games app E
Your method counts these two selections as different (the italicised part is the "at least one of each app" requirement), but they are the same.
answered 21 mins ago
Parcly Taxel
35.6k136991
Ok, I see now why my answer is wrong but, can you point out where/what step in my solution this overlap occurs? Is it possible to change my solution to avoid this error?
– abnas
5 mins ago
@abnas When you start saying "the remaining two apps are selected from the unselected 20", that is where you overcount. There is no easy fix to your attempt.
– Parcly Taxel
4 mins ago
add a comment |Â
up vote
3
down vote
You are overcounting many, many choices, such as the following selections:
train app A, SNS app B, games app C, games app D, games app E
train app A, SNS app B, games app D, games app C, games app E
Your method counts these two selections as different (the italicised part is the "at least one of each app" requirement), but they are the same.
answered 21 mins ago
Parcly Taxel
35.6k136991
Ok, I see now why my answer is wrong but, can you point out where/what step in my solution this overlap occurs? Is it possible to change my solution to avoid this error?
– abnas
5 mins ago
@abnas When you start saying "the remaining two apps are selected from the unselected 20", that is where you overcount. There is no easy fix to your attempt.
– Parcly Taxel
4 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You are overcounting many, many choices, such as the following selections:
train app A, SNS app B, games app C, games app D, games app E
train app A, SNS app B, games app D, games app C, games app E
Your method counts these two selections as different (the italicised part is the "at least one of each app" requirement), but they are the same.
answered 21 mins ago
Parcly Taxel
35.6k136991
You are overcounting many, many choices, such as the following selections:
train app A, SNS app B, games app C, games app D, games app E
train app A, SNS app B, games app D, games app C, games app E
Your method counts these two selections as different (the italicised part is the "at least one of each app" requirement), but they are the same.
answered 21 mins ago
Parcly Taxel
35.6k136991
answered 21 mins ago
Parcly Taxel
35.6k136991
answered 21 mins ago
Parcly Taxel
35.6k136991
answered 21 mins ago
Parcly Taxel
35.6k136991
35.6k136991
Ok, I see now why my answer is wrong but, can you point out where/what step in my solution this overlap occurs? Is it possible to change my solution to avoid this error?
– abnas
5 mins ago
@abnas When you start saying "the remaining two apps are selected from the unselected 20", that is where you overcount. There is no easy fix to your attempt.
– Parcly Taxel
4 mins ago
add a comment |Â
Ok, I see now why my answer is wrong but, can you point out where/what step in my solution this overlap occurs? Is it possible to change my solution to avoid this error?
– abnas
5 mins ago
@abnas When you start saying "the remaining two apps are selected from the unselected 20", that is where you overcount. There is no easy fix to your attempt.
– Parcly Taxel
4 mins ago
Ok, I see now why my answer is wrong but, can you point out where/what step in my solution this overlap occurs? Is it possible to change my solution to avoid this error?
– abnas
5 mins ago
Ok, I see now why my answer is wrong but, can you point out where/what step in my solution this overlap occurs? Is it possible to change my solution to avoid this error?
– abnas
5 mins ago
@abnas When you start saying "the remaining two apps are selected from the unselected 20", that is where you overcount. There is no easy fix to your attempt.
– Parcly Taxel
4 mins ago
@abnas When you start saying "the remaining two apps are selected from the unselected 20", that is where you overcount. There is no easy fix to your attempt.
– Parcly Taxel
4 mins ago
add a comment |Â
up vote
0
down vote
As pointed out by @Parcly Taxel,
You are overcounting the apps; keep in mind that order is irrelevant here, so train1,train2 is same as train2,train1. This is where overccounting crept into your solution.
So, how to avoid this overcounting?
Take 3T,6S and 14G apps.
In total, you want 5 apps.
So make selections like: $$(1T,1S,3G),(1T,3S,1G),(3T,1S,1G),(2T,2S,1G),(2T,1S,2G),(1T,2S,2G)$$
Now, count for number of ways you have for each of the above selection, and add.
answered 4 mins ago
omega
8042818
add a comment |Â
up vote
0
down vote
As pointed out by @Parcly Taxel,
You are overcounting the apps; keep in mind that order is irrelevant here, so train1,train2 is same as train2,train1. This is where overccounting crept into your solution.
So, how to avoid this overcounting?
Take 3T,6S and 14G apps.
In total, you want 5 apps.
So make selections like: $$(1T,1S,3G),(1T,3S,1G),(3T,1S,1G),(2T,2S,1G),(2T,1S,2G),(1T,2S,2G)$$
Now, count for number of ways you have for each of the above selection, and add.
answered 4 mins ago
omega
8042818
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As pointed out by @Parcly Taxel,
You are overcounting the apps; keep in mind that order is irrelevant here, so train1,train2 is same as train2,train1. This is where overccounting crept into your solution.
So, how to avoid this overcounting?
Take 3T,6S and 14G apps.
In total, you want 5 apps.
So make selections like: $$(1T,1S,3G),(1T,3S,1G),(3T,1S,1G),(2T,2S,1G),(2T,1S,2G),(1T,2S,2G)$$
Now, count for number of ways you have for each of the above selection, and add.
answered 4 mins ago
omega
8042818
As pointed out by @Parcly Taxel,
You are overcounting the apps; keep in mind that order is irrelevant here, so train1,train2 is same as train2,train1. This is where overccounting crept into your solution.
So, how to avoid this overcounting?
Take 3T,6S and 14G apps.
In total, you want 5 apps.
So make selections like: $$(1T,1S,3G),(1T,3S,1G),(3T,1S,1G),(2T,2S,1G),(2T,1S,2G),(1T,2S,2G)$$
Now, count for number of ways you have for each of the above selection, and add.
answered 4 mins ago
omega
8042818
answered 4 mins ago
omega
8042818
answered 4 mins ago
omega
8042818
answered 4 mins ago
omega
8042818
8042818
add a comment |Â
add a comment |Â
up vote
0
down vote
As Parcly Taxel pointed out, you overcount many times because picking the people in two different counts(using your method) doesn't account for the ordering.
My attempt(casework) is the following:
The possible group sizes per element must be either: 1, 1, 3 or 1, 2, 2.
Case 1: 1, 1, 3
There are different combinations based on the different choices of the largest group. So we compute them separately, giving
$$binom31binom61binom143 + binom31binom63binom141 + binom33binom61binom141 = 7476.$$
Case 2: 1, 2, 2
Similarly the the last one, we have
$$binom31binom62binom142+binom32binom61binom142+binom32binom62binom141 = 6363.$$
Summing the two cases gives a total of $$boxed13839.$$
answered 2 mins ago
math783625
1037
add a comment |Â
up vote
0
down vote
As Parcly Taxel pointed out, you overcount many times because picking the people in two different counts(using your method) doesn't account for the ordering.
My attempt(casework) is the following:
The possible group sizes per element must be either: 1, 1, 3 or 1, 2, 2.
Case 1: 1, 1, 3
There are different combinations based on the different choices of the largest group. So we compute them separately, giving
$$binom31binom61binom143 + binom31binom63binom141 + binom33binom61binom141 = 7476.$$
Case 2: 1, 2, 2
Similarly the the last one, we have
$$binom31binom62binom142+binom32binom61binom142+binom32binom62binom141 = 6363.$$
Summing the two cases gives a total of $$boxed13839.$$
answered 2 mins ago
math783625
1037
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As Parcly Taxel pointed out, you overcount many times because picking the people in two different counts(using your method) doesn't account for the ordering.
My attempt(casework) is the following:
The possible group sizes per element must be either: 1, 1, 3 or 1, 2, 2.
Case 1: 1, 1, 3
There are different combinations based on the different choices of the largest group. So we compute them separately, giving
$$binom31binom61binom143 + binom31binom63binom141 + binom33binom61binom141 = 7476.$$
Case 2: 1, 2, 2
Similarly the the last one, we have
$$binom31binom62binom142+binom32binom61binom142+binom32binom62binom141 = 6363.$$
Summing the two cases gives a total of $$boxed13839.$$
answered 2 mins ago
math783625
1037
As Parcly Taxel pointed out, you overcount many times because picking the people in two different counts(using your method) doesn't account for the ordering.
My attempt(casework) is the following:
The possible group sizes per element must be either: 1, 1, 3 or 1, 2, 2.
Case 1: 1, 1, 3
There are different combinations based on the different choices of the largest group. So we compute them separately, giving
$$binom31binom61binom143 + binom31binom63binom141 + binom33binom61binom141 = 7476.$$
Case 2: 1, 2, 2
Similarly the the last one, we have
$$binom31binom62binom142+binom32binom61binom142+binom32binom62binom141 = 6363.$$
Summing the two cases gives a total of $$boxed13839.$$
answered 2 mins ago
math783625
1037
answered 2 mins ago
math783625
1037
answered 2 mins ago
math783625
1037
answered 2 mins ago
math783625
1037
1037
add a comment |Â
add a comment |Â
abnas is a new contributor. Be nice, and check out our Code of Conduct.
abnas is a new contributor. Be nice, and check out our Code of Conduct.
abnas is a new contributor. Be nice, and check out our Code of Conduct.
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