Mixing
Index summation over a binomial coefficiant series [closed]
-2
$begingroup$
Can the following sum be reduced to an analytic expression?
$$sum_{m=0}^Nfrac{N!}{m!(N-m)!}$$
summation binomial-coefficients
asked Jan 27 at 16:14
jarheadjarhead
1308
$endgroup$
closed as off-topic by callculus, NCh, Andrew, max_zorn, saz Jan 30 at 7:52
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." – callculus, NCh, Andrew, max_zorn, saz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
-2
$begingroup$
Can the following sum be reduced to an analytic expression?
$$sum_{m=0}^Nfrac{N!}{m!(N-m)!}$$
summation binomial-coefficients
asked Jan 27 at 16:14
jarheadjarhead
1308
$endgroup$
closed as off-topic by callculus, NCh, Andrew, max_zorn, saz Jan 30 at 7:52
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." – callculus, NCh, Andrew, max_zorn, saz
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
$endgroup$
– callculus
Jan 27 at 16:20
2
$begingroup$
This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
$endgroup$
– user376343
Jan 27 at 16:24
$begingroup$
For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
$endgroup$
– Ethan Bolker
Jan 27 at 16:37
add a comment |
-2
-2
-2
$begingroup$
Can the following sum be reduced to an analytic expression?
$$sum_{m=0}^Nfrac{N!}{m!(N-m)!}$$
summation binomial-coefficients
asked Jan 27 at 16:14
jarheadjarhead
1308
$endgroup$
Can the following sum be reduced to an analytic expression?
$$sum_{m=0}^Nfrac{N!}{m!(N-m)!}$$
summation binomial-coefficients
summation binomial-coefficients
asked Jan 27 at 16:14
jarheadjarhead
1308
asked Jan 27 at 16:14
jarheadjarhead
1308
asked Jan 27 at 16:14
jarheadjarhead
1308
asked Jan 27 at 16:14
jarheadjarhead
1308
asked Jan 27 at 16:14
jarheadjarhead
1308
1308
closed as off-topic by callculus, NCh, Andrew, max_zorn, saz Jan 30 at 7:52
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." – callculus, NCh, Andrew, max_zorn, saz
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by callculus, NCh, Andrew, max_zorn, saz Jan 30 at 7:52
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." – callculus, NCh, Andrew, max_zorn, saz
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
$endgroup$
– callculus
Jan 27 at 16:20
2
$begingroup$
This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
$endgroup$
– user376343
Jan 27 at 16:24
$begingroup$
For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
$endgroup$
– Ethan Bolker
Jan 27 at 16:37
add a comment |
1
$begingroup$
Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
$endgroup$
– callculus
Jan 27 at 16:20
2
$begingroup$
This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
$endgroup$
– user376343
Jan 27 at 16:24
$begingroup$
For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
$endgroup$
– Ethan Bolker
Jan 27 at 16:37
1
1
$begingroup$
Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
$endgroup$
– callculus
Jan 27 at 16:20
$begingroup$
Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
$endgroup$
– callculus
Jan 27 at 16:20
2
2
$begingroup$
This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
$endgroup$
– user376343
Jan 27 at 16:24
$begingroup$
This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
$endgroup$
– user376343
Jan 27 at 16:24
$begingroup$
For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
$endgroup$
– Ethan Bolker
Jan 27 at 16:37
$begingroup$
For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
$endgroup$
– Ethan Bolker
Jan 27 at 16:37
add a comment |
1 Answer
1
active
oldest
votes
1
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Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$
Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$
answered Jan 27 at 16:32
Exp ikxExp ikx
4489
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$
Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$
answered Jan 27 at 16:32
Exp ikxExp ikx
4489
$endgroup$
add a comment |
1
$begingroup$
Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$
Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$
answered Jan 27 at 16:32
Exp ikxExp ikx
4489
$endgroup$
add a comment |
1
1
1
$begingroup$
Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$
Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$
answered Jan 27 at 16:32
Exp ikxExp ikx
4489
$endgroup$
Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$
Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$
answered Jan 27 at 16:32
Exp ikxExp ikx
4489
answered Jan 27 at 16:32
Exp ikxExp ikx
4489
answered Jan 27 at 16:32
Exp ikxExp ikx
4489
answered Jan 27 at 16:32
Exp ikxExp ikx
4489
4489
add a comment |
add a comment |
1
$begingroup$
Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
$endgroup$
– callculus
Jan 27 at 16:20
2
$begingroup$
This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
$endgroup$
– user376343
Jan 27 at 16:24
$begingroup$
For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
$endgroup$
– Ethan Bolker
Jan 27 at 16:37