The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all...
$begingroup$
Question based on Permutations and Combinations
The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all different is ?
A) $2^{20} + binom{20}{10}$
B) $2^{20} - binom{20}{9}$
C) $2^{19} + binom{20}{9}$
D) $2^{19} - binom{20}{10}$
combinatorics permutations combinations
edited Jan 5 at 5:55
dmtri
1,4521521
asked Jan 5 at 5:04
Vysakh AVVysakh AV
94
New contributor
Vysakh AV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as off-topic by Saad, user91500, mrtaurho, Kenny Wong, darij grinberg Jan 5 at 14:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, user91500, mrtaurho, Kenny Wong, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Question based on Permutations and Combinations
The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all different is ?
A) $2^{20} + binom{20}{10}$
B) $2^{20} - binom{20}{9}$
C) $2^{19} + binom{20}{9}$
D) $2^{19} - binom{20}{10}$
combinatorics permutations combinations
edited Jan 5 at 5:55
dmtri
1,4521521
asked Jan 5 at 5:04
Vysakh AVVysakh AV
94
New contributor
Vysakh AV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as off-topic by Saad, user91500, mrtaurho, Kenny Wong, darij grinberg Jan 5 at 14:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, user91500, mrtaurho, Kenny Wong, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Then you essentially only have $21$ objects to choose from, in which case the answer would be $C^{21}_{10}$
$endgroup$
– glowstonetrees
Jan 5 at 5:08
4
$begingroup$
Nope cause you can choose like 5 of the alike things and then 5 distinct things and such combinations
$endgroup$
– Vysakh AV
Jan 5 at 5:12
add a comment |
$begingroup$
Question based on Permutations and Combinations
The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all different is ?
A) $2^{20} + binom{20}{10}$
B) $2^{20} - binom{20}{9}$
C) $2^{19} + binom{20}{9}$
D) $2^{19} - binom{20}{10}$
combinatorics permutations combinations
edited Jan 5 at 5:55
dmtri
1,4521521
asked Jan 5 at 5:04
Vysakh AVVysakh AV
94
New contributor
Vysakh AV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Question based on Permutations and Combinations
The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all different is ?
A) $2^{20} + binom{20}{10}$
B) $2^{20} - binom{20}{9}$
C) $2^{19} + binom{20}{9}$
D) $2^{19} - binom{20}{10}$
combinatorics permutations combinations
combinatorics permutations combinations
edited Jan 5 at 5:55
dmtri
1,4521521
asked Jan 5 at 5:04
Vysakh AVVysakh AV
94
New contributor
Vysakh AV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 5 at 5:55
dmtri
1,4521521
asked Jan 5 at 5:04
Vysakh AVVysakh AV
94
New contributor
Vysakh AV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 5 at 5:55
dmtri
1,4521521
edited Jan 5 at 5:55
dmtri
1,4521521
edited Jan 5 at 5:55
dmtri
1,4521521
1,4521521
asked Jan 5 at 5:04
Vysakh AVVysakh AV
94
New contributor
Vysakh AV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 5 at 5:04
Vysakh AVVysakh AV
94
asked Jan 5 at 5:04
Vysakh AVVysakh AV
94
94
New contributor
Vysakh AV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Vysakh AV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Vysakh AV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as off-topic by Saad, user91500, mrtaurho, Kenny Wong, darij grinberg Jan 5 at 14:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, user91500, mrtaurho, Kenny Wong, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, user91500, mrtaurho, Kenny Wong, darij grinberg Jan 5 at 14:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, user91500, mrtaurho, Kenny Wong, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Then you essentially only have $21$ objects to choose from, in which case the answer would be $C^{21}_{10}$
$endgroup$
– glowstonetrees
Jan 5 at 5:08
4
$begingroup$
Nope cause you can choose like 5 of the alike things and then 5 distinct things and such combinations
$endgroup$
– Vysakh AV
Jan 5 at 5:12
add a comment |
1
$begingroup$
Then you essentially only have $21$ objects to choose from, in which case the answer would be $C^{21}_{10}$
$endgroup$
– glowstonetrees
Jan 5 at 5:08
4
$begingroup$
Nope cause you can choose like 5 of the alike things and then 5 distinct things and such combinations
$endgroup$
– Vysakh AV
Jan 5 at 5:12
1
1
$begingroup$
Then you essentially only have $21$ objects to choose from, in which case the answer would be $C^{21}_{10}$
$endgroup$
– glowstonetrees
Jan 5 at 5:08
$begingroup$
Then you essentially only have $21$ objects to choose from, in which case the answer would be $C^{21}_{10}$
$endgroup$
– glowstonetrees
Jan 5 at 5:08
4
4
$begingroup$
Nope cause you can choose like 5 of the alike things and then 5 distinct things and such combinations
$endgroup$
– Vysakh AV
Jan 5 at 5:12
$begingroup$
Nope cause you can choose like 5 of the alike things and then 5 distinct things and such combinations
$endgroup$
– Vysakh AV
Jan 5 at 5:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you choose $n$ of the unique object, then there are $20choose n$ ways to pick those, and $1$ to pick the other $10-n$ non-unique ones. So the answer is
$$sum_{n=0}^{10} {20choose n}$$
By symmetry, we have
$$2sum_{n=0}^{10} {20choose n} = sum_{n=0}^{20} {20choose n} + {20choose 10} = 2^{20} + {20choose 10}$$
So
$$sum_{n=0}^{10} {20choose n} = 2^{19} + frac12{20choose 10}$$
And ${19choose10} = {19choose9}$ and ${19choose10} + {19choose9} = {20choose10}$ by Pascal's Identity, so this gives $frac12 {20choose10} = {19choose9}$ so the final answer is
$$2^{19} + {19choose 9}$$
answered Jan 5 at 5:24
Erik ParkinsonErik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Suppose you pick $p$ objects from the set $P$ of identical objects, and let the set $Q$ be that of the other $20$ objects. Scaling $p$ in the range $[0, 10]$ and permuting the ways to complete the $10$ with $Q$ gives us:
$$sum_{p=0}^{10}binom{20}{10-p}$$
answered Jan 5 at 5:24
Rhys HughesRhys Hughes
5,1501427
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you choose $n$ of the unique object, then there are $20choose n$ ways to pick those, and $1$ to pick the other $10-n$ non-unique ones. So the answer is
$$sum_{n=0}^{10} {20choose n}$$
By symmetry, we have
$$2sum_{n=0}^{10} {20choose n} = sum_{n=0}^{20} {20choose n} + {20choose 10} = 2^{20} + {20choose 10}$$
So
$$sum_{n=0}^{10} {20choose n} = 2^{19} + frac12{20choose 10}$$
And ${19choose10} = {19choose9}$ and ${19choose10} + {19choose9} = {20choose10}$ by Pascal's Identity, so this gives $frac12 {20choose10} = {19choose9}$ so the final answer is
$$2^{19} + {19choose 9}$$
answered Jan 5 at 5:24
Erik ParkinsonErik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If you choose $n$ of the unique object, then there are $20choose n$ ways to pick those, and $1$ to pick the other $10-n$ non-unique ones. So the answer is
$$sum_{n=0}^{10} {20choose n}$$
By symmetry, we have
$$2sum_{n=0}^{10} {20choose n} = sum_{n=0}^{20} {20choose n} + {20choose 10} = 2^{20} + {20choose 10}$$
So
$$sum_{n=0}^{10} {20choose n} = 2^{19} + frac12{20choose 10}$$
And ${19choose10} = {19choose9}$ and ${19choose10} + {19choose9} = {20choose10}$ by Pascal's Identity, so this gives $frac12 {20choose10} = {19choose9}$ so the final answer is
$$2^{19} + {19choose 9}$$
answered Jan 5 at 5:24
Erik ParkinsonErik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If you choose $n$ of the unique object, then there are $20choose n$ ways to pick those, and $1$ to pick the other $10-n$ non-unique ones. So the answer is
$$sum_{n=0}^{10} {20choose n}$$
By symmetry, we have
$$2sum_{n=0}^{10} {20choose n} = sum_{n=0}^{20} {20choose n} + {20choose 10} = 2^{20} + {20choose 10}$$
So
$$sum_{n=0}^{10} {20choose n} = 2^{19} + frac12{20choose 10}$$
And ${19choose10} = {19choose9}$ and ${19choose10} + {19choose9} = {20choose10}$ by Pascal's Identity, so this gives $frac12 {20choose10} = {19choose9}$ so the final answer is
$$2^{19} + {19choose 9}$$
answered Jan 5 at 5:24
Erik ParkinsonErik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If you choose $n$ of the unique object, then there are $20choose n$ ways to pick those, and $1$ to pick the other $10-n$ non-unique ones. So the answer is
$$sum_{n=0}^{10} {20choose n}$$
By symmetry, we have
$$2sum_{n=0}^{10} {20choose n} = sum_{n=0}^{20} {20choose n} + {20choose 10} = 2^{20} + {20choose 10}$$
So
$$sum_{n=0}^{10} {20choose n} = 2^{19} + frac12{20choose 10}$$
And ${19choose10} = {19choose9}$ and ${19choose10} + {19choose9} = {20choose10}$ by Pascal's Identity, so this gives $frac12 {20choose10} = {19choose9}$ so the final answer is
$$2^{19} + {19choose 9}$$
answered Jan 5 at 5:24
Erik ParkinsonErik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 5 at 5:24
Erik ParkinsonErik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 5 at 5:24
Erik ParkinsonErik Parkinson
9159
answered Jan 5 at 5:24
Erik ParkinsonErik Parkinson
9159
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Erik Parkinson 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 |
add a comment |
$begingroup$
Suppose you pick $p$ objects from the set $P$ of identical objects, and let the set $Q$ be that of the other $20$ objects. Scaling $p$ in the range $[0, 10]$ and permuting the ways to complete the $10$ with $Q$ gives us:
$$sum_{p=0}^{10}binom{20}{10-p}$$
answered Jan 5 at 5:24
Rhys HughesRhys Hughes
5,1501427
$endgroup$
add a comment |
$begingroup$
Suppose you pick $p$ objects from the set $P$ of identical objects, and let the set $Q$ be that of the other $20$ objects. Scaling $p$ in the range $[0, 10]$ and permuting the ways to complete the $10$ with $Q$ gives us:
$$sum_{p=0}^{10}binom{20}{10-p}$$
answered Jan 5 at 5:24
Rhys HughesRhys Hughes
5,1501427
$endgroup$
add a comment |
$begingroup$
Suppose you pick $p$ objects from the set $P$ of identical objects, and let the set $Q$ be that of the other $20$ objects. Scaling $p$ in the range $[0, 10]$ and permuting the ways to complete the $10$ with $Q$ gives us:
$$sum_{p=0}^{10}binom{20}{10-p}$$
answered Jan 5 at 5:24
Rhys HughesRhys Hughes
5,1501427
$endgroup$
Suppose you pick $p$ objects from the set $P$ of identical objects, and let the set $Q$ be that of the other $20$ objects. Scaling $p$ in the range $[0, 10]$ and permuting the ways to complete the $10$ with $Q$ gives us:
$$sum_{p=0}^{10}binom{20}{10-p}$$
answered Jan 5 at 5:24
Rhys HughesRhys Hughes
5,1501427
answered Jan 5 at 5:24
Rhys HughesRhys Hughes
5,1501427
answered Jan 5 at 5:24
Rhys HughesRhys Hughes
5,1501427
answered Jan 5 at 5:24
Rhys HughesRhys Hughes
5,1501427
5,1501427
add a comment |
add a comment |
1
$begingroup$
Then you essentially only have $21$ objects to choose from, in which case the answer would be $C^{21}_{10}$
$endgroup$
– glowstonetrees
Jan 5 at 5:08
4
$begingroup$
Nope cause you can choose like 5 of the alike things and then 5 distinct things and such combinations
$endgroup$
– Vysakh AV
Jan 5 at 5:12