Serge Lang - Basic Mathematics - Sum of Binomial coefficients p.387 ex. 9
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I can not understand the solution.How does the second step come about? How does the expression in brackets appear?
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binomial-coefficients
asked Nov 20 at 8:13
Alexander Nikolin
51
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up vote
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favorite
I can not understand the solution.How does the second step come about? How does the expression in brackets appear?
solution image
binomial-coefficients
asked Nov 20 at 8:13
Alexander Nikolin
51
$ab + ac = a[b+c]$ :-)
– trancelocation
Nov 20 at 8:16
yeah, I know, but here it is somehow not obvious to me
– Alexander Nikolin
Nov 20 at 8:20
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I can not understand the solution.How does the second step come about? How does the expression in brackets appear?
solution image
binomial-coefficients
asked Nov 20 at 8:13
Alexander Nikolin
51
I can not understand the solution.How does the second step come about? How does the expression in brackets appear?
solution image
binomial-coefficients
binomial-coefficients
asked Nov 20 at 8:13
Alexander Nikolin
51
asked Nov 20 at 8:13
Alexander Nikolin
51
asked Nov 20 at 8:13
Alexander Nikolin
51
asked Nov 20 at 8:13
Alexander Nikolin
51
asked Nov 20 at 8:13
Alexander Nikolin
51
51
$ab + ac = a[b+c]$ :-)
– trancelocation
Nov 20 at 8:16
yeah, I know, but here it is somehow not obvious to me
– Alexander Nikolin
Nov 20 at 8:20
add a comment |
$ab + ac = a[b+c]$ :-)
– trancelocation
Nov 20 at 8:16
yeah, I know, but here it is somehow not obvious to me
– Alexander Nikolin
Nov 20 at 8:20
$ab + ac = a[b+c]$ :-)
– trancelocation
Nov 20 at 8:16
$ab + ac = a[b+c]$ :-)
– trancelocation
Nov 20 at 8:16
yeah, I know, but here it is somehow not obvious to me
– Alexander Nikolin
Nov 20 at 8:20
yeah, I know, but here it is somehow not obvious to me
– Alexander Nikolin
Nov 20 at 8:20
add a comment |
1 Answer
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1
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Take note:
$$(n - k + 1)! = (n-k)! cdot (n - k + 1)$$
$$k! = (k-1)! cdot k$$
Rewrite the expression after the first equal sign with this in mind: you should see a common factor you can neatly factor out.
answered Nov 20 at 8:17
Eevee Trainer
2,812222
thanks! the problem was that the first property was not clear to me
– Alexander Nikolin
Nov 20 at 8:25
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Take note:
$$(n - k + 1)! = (n-k)! cdot (n - k + 1)$$
$$k! = (k-1)! cdot k$$
Rewrite the expression after the first equal sign with this in mind: you should see a common factor you can neatly factor out.
answered Nov 20 at 8:17
Eevee Trainer
2,812222
thanks! the problem was that the first property was not clear to me
– Alexander Nikolin
Nov 20 at 8:25
add a comment |
up vote
1
down vote
accepted
Take note:
$$(n - k + 1)! = (n-k)! cdot (n - k + 1)$$
$$k! = (k-1)! cdot k$$
Rewrite the expression after the first equal sign with this in mind: you should see a common factor you can neatly factor out.
answered Nov 20 at 8:17
Eevee Trainer
2,812222
thanks! the problem was that the first property was not clear to me
– Alexander Nikolin
Nov 20 at 8:25
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Take note:
$$(n - k + 1)! = (n-k)! cdot (n - k + 1)$$
$$k! = (k-1)! cdot k$$
Rewrite the expression after the first equal sign with this in mind: you should see a common factor you can neatly factor out.
answered Nov 20 at 8:17
Eevee Trainer
2,812222
Take note:
$$(n - k + 1)! = (n-k)! cdot (n - k + 1)$$
$$k! = (k-1)! cdot k$$
Rewrite the expression after the first equal sign with this in mind: you should see a common factor you can neatly factor out.
answered Nov 20 at 8:17
Eevee Trainer
2,812222
answered Nov 20 at 8:17
Eevee Trainer
2,812222
answered Nov 20 at 8:17
Eevee Trainer
2,812222
answered Nov 20 at 8:17
Eevee Trainer
2,812222
2,812222
thanks! the problem was that the first property was not clear to me
– Alexander Nikolin
Nov 20 at 8:25
add a comment |
thanks! the problem was that the first property was not clear to me
– Alexander Nikolin
Nov 20 at 8:25
thanks! the problem was that the first property was not clear to me
– Alexander Nikolin
Nov 20 at 8:25
thanks! the problem was that the first property was not clear to me
– Alexander Nikolin
Nov 20 at 8:25
add a comment |
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$ab + ac = a[b+c]$ :-)
– trancelocation
Nov 20 at 8:16
yeah, I know, but here it is somehow not obvious to me
– Alexander Nikolin
Nov 20 at 8:20