The sum $sum_{r=0}^{20}r(20-r)binom{20}{r}^2$
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Find the value of: $$displaystyle sum_{r=0}^{20}r(20-r)dbinom{20}{r}^2$$
Attempt:
Using, $dbinom{n}{r}= dfrac n r dbinom{n-1}{r-1}$, I get $400sum_{r=0}^{20}left(dbinom {19} r dbinom{19}{r-1}right)$
What do I do next? I tried to use $dbinom n r +dbinom n {r-1}= dbinom{n+1}{r}$ but that didn't help. Also, I know that the sum of square of binomial coefficients is $dbinom {2n}{n}$ and the Vandermonde's identity is $dbinom{n+m}{r}= sum _{k=0}^r dbinom m k dbinom n {r-k}$.
combinatorics summation binomial-coefficients binomial-theorem
edited Nov 16 at 11:51
Martin Sleziak
44.4k7115268
asked Feb 28 at 21:08
Abcd
2,88911130
add a comment |
up vote
2
down vote
favorite
Find the value of: $$displaystyle sum_{r=0}^{20}r(20-r)dbinom{20}{r}^2$$
Attempt:
Using, $dbinom{n}{r}= dfrac n r dbinom{n-1}{r-1}$, I get $400sum_{r=0}^{20}left(dbinom {19} r dbinom{19}{r-1}right)$
What do I do next? I tried to use $dbinom n r +dbinom n {r-1}= dbinom{n+1}{r}$ but that didn't help. Also, I know that the sum of square of binomial coefficients is $dbinom {2n}{n}$ and the Vandermonde's identity is $dbinom{n+m}{r}= sum _{k=0}^r dbinom m k dbinom n {r-k}$.
combinatorics summation binomial-coefficients binomial-theorem
edited Nov 16 at 11:51
Martin Sleziak
44.4k7115268
asked Feb 28 at 21:08
Abcd
2,88911130
1
Then you replace $binom{19}{r-1}$ by $binom{19}{20-r}$, and think about some combinatorial arguments comparing your summation and $binom{19+19}{r+(20-r)} = binom{38}{20}$... (You have 19 black balls and 19 white balls, in how many ways can you pick 20 balls...?)
– Hw Chu
Feb 28 at 21:14
@HwChu Is it equivalent to the number of non negative integer solutions of $x+y = 20$?
– Abcd
Feb 28 at 21:25
@HwChu I'd like to see the combinatorial solution
– Abcd
Feb 28 at 21:30
1
I wanted to say this is just the Vandermonde's identity. I did not know that identity had a name when I made the comment, so you may just look at the combinarotial proof of Vandermonde's identity.
– Hw Chu
Feb 28 at 21:45
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find the value of: $$displaystyle sum_{r=0}^{20}r(20-r)dbinom{20}{r}^2$$
Attempt:
Using, $dbinom{n}{r}= dfrac n r dbinom{n-1}{r-1}$, I get $400sum_{r=0}^{20}left(dbinom {19} r dbinom{19}{r-1}right)$
What do I do next? I tried to use $dbinom n r +dbinom n {r-1}= dbinom{n+1}{r}$ but that didn't help. Also, I know that the sum of square of binomial coefficients is $dbinom {2n}{n}$ and the Vandermonde's identity is $dbinom{n+m}{r}= sum _{k=0}^r dbinom m k dbinom n {r-k}$.
combinatorics summation binomial-coefficients binomial-theorem
edited Nov 16 at 11:51
Martin Sleziak
44.4k7115268
asked Feb 28 at 21:08
Abcd
2,88911130
Find the value of: $$displaystyle sum_{r=0}^{20}r(20-r)dbinom{20}{r}^2$$
Attempt:
Using, $dbinom{n}{r}= dfrac n r dbinom{n-1}{r-1}$, I get $400sum_{r=0}^{20}left(dbinom {19} r dbinom{19}{r-1}right)$
What do I do next? I tried to use $dbinom n r +dbinom n {r-1}= dbinom{n+1}{r}$ but that didn't help. Also, I know that the sum of square of binomial coefficients is $dbinom {2n}{n}$ and the Vandermonde's identity is $dbinom{n+m}{r}= sum _{k=0}^r dbinom m k dbinom n {r-k}$.
combinatorics summation binomial-coefficients binomial-theorem
combinatorics summation binomial-coefficients binomial-theorem
edited Nov 16 at 11:51
Martin Sleziak
44.4k7115268
asked Feb 28 at 21:08
Abcd
2,88911130
edited Nov 16 at 11:51
Martin Sleziak
44.4k7115268
asked Feb 28 at 21:08
Abcd
2,88911130
edited Nov 16 at 11:51
Martin Sleziak
44.4k7115268
edited Nov 16 at 11:51
Martin Sleziak
44.4k7115268
edited Nov 16 at 11:51
Martin Sleziak
44.4k7115268
44.4k7115268
asked Feb 28 at 21:08
Abcd
2,88911130
asked Feb 28 at 21:08
Abcd
2,88911130
asked Feb 28 at 21:08
Abcd
2,88911130
2,88911130
1
Then you replace $binom{19}{r-1}$ by $binom{19}{20-r}$, and think about some combinatorial arguments comparing your summation and $binom{19+19}{r+(20-r)} = binom{38}{20}$... (You have 19 black balls and 19 white balls, in how many ways can you pick 20 balls...?)
– Hw Chu
Feb 28 at 21:14
@HwChu Is it equivalent to the number of non negative integer solutions of $x+y = 20$?
– Abcd
Feb 28 at 21:25
@HwChu I'd like to see the combinatorial solution
– Abcd
Feb 28 at 21:30
1
I wanted to say this is just the Vandermonde's identity. I did not know that identity had a name when I made the comment, so you may just look at the combinarotial proof of Vandermonde's identity.
– Hw Chu
Feb 28 at 21:45
add a comment |
1
Then you replace $binom{19}{r-1}$ by $binom{19}{20-r}$, and think about some combinatorial arguments comparing your summation and $binom{19+19}{r+(20-r)} = binom{38}{20}$... (You have 19 black balls and 19 white balls, in how many ways can you pick 20 balls...?)
– Hw Chu
Feb 28 at 21:14
@HwChu Is it equivalent to the number of non negative integer solutions of $x+y = 20$?
– Abcd
Feb 28 at 21:25
@HwChu I'd like to see the combinatorial solution
– Abcd
Feb 28 at 21:30
1
I wanted to say this is just the Vandermonde's identity. I did not know that identity had a name when I made the comment, so you may just look at the combinarotial proof of Vandermonde's identity.
– Hw Chu
Feb 28 at 21:45
1
1
Then you replace $binom{19}{r-1}$ by $binom{19}{20-r}$, and think about some combinatorial arguments comparing your summation and $binom{19+19}{r+(20-r)} = binom{38}{20}$... (You have 19 black balls and 19 white balls, in how many ways can you pick 20 balls...?)
– Hw Chu
Feb 28 at 21:14
Then you replace $binom{19}{r-1}$ by $binom{19}{20-r}$, and think about some combinatorial arguments comparing your summation and $binom{19+19}{r+(20-r)} = binom{38}{20}$... (You have 19 black balls and 19 white balls, in how many ways can you pick 20 balls...?)
– Hw Chu
Feb 28 at 21:14
@HwChu Is it equivalent to the number of non negative integer solutions of $x+y = 20$?
– Abcd
Feb 28 at 21:25
@HwChu Is it equivalent to the number of non negative integer solutions of $x+y = 20$?
– Abcd
Feb 28 at 21:25
@HwChu I'd like to see the combinatorial solution
– Abcd
Feb 28 at 21:30
@HwChu I'd like to see the combinatorial solution
– Abcd
Feb 28 at 21:30
1
1
I wanted to say this is just the Vandermonde's identity. I did not know that identity had a name when I made the comment, so you may just look at the combinarotial proof of Vandermonde's identity.
– Hw Chu
Feb 28 at 21:45
I wanted to say this is just the Vandermonde's identity. I did not know that identity had a name when I made the comment, so you may just look at the combinarotial proof of Vandermonde's identity.
– Hw Chu
Feb 28 at 21:45
add a comment |
2 Answers
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2
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Hint: Write the sum as $$sum_{r=0}^{20}left(rbinom{20}{r}right)cdotleft((20-r)binom{20}{20-r}right) = sum_{r = 0}^{20}a_ra_{20-r},$$
where $a_r = rbinom{20}{r}.$ This is the coefficient of $x^{20}$ in the expansion of $$left(sum_{r=0}^{20}a_rx^rright)^2.$$
answered Feb 28 at 21:21
dezdichado
5,7391928
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$$sum_{r=0}^{20} r(20-r)binom {20}{r} ^2=20sum_{r=0}^{20} rbinom {20}{r} ^2- sum_{r=0}^{20} r^2binom {20}{r} ^2=400sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - 400sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}$$
Hence by Vandermonde's identity we get $$400left(sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}right) =400left(binom {39}{19}- binom {38}{19}right) $$
Now by Pascal's rule $$400left(binom {39}{19}- binom {38}{19}right) =400binom {38}{18}$$
Alternatively you can reach answer by your way too
$$400sum_{color{red}{r=1}}^{color{red}{19}} binom {19}{r}binom {19}{20-r}$$
Hence by using Vandermonde's identity we get $$400sum_{r=1}^{19} binom {19}{r}binom {19}{20-r}=400binom {38}{20}=400binom {38}{18}$$
answered Mar 7 at 16:51
Digamma
5,8241437
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint: Write the sum as $$sum_{r=0}^{20}left(rbinom{20}{r}right)cdotleft((20-r)binom{20}{20-r}right) = sum_{r = 0}^{20}a_ra_{20-r},$$
where $a_r = rbinom{20}{r}.$ This is the coefficient of $x^{20}$ in the expansion of $$left(sum_{r=0}^{20}a_rx^rright)^2.$$
answered Feb 28 at 21:21
dezdichado
5,7391928
add a comment |
up vote
2
down vote
Hint: Write the sum as $$sum_{r=0}^{20}left(rbinom{20}{r}right)cdotleft((20-r)binom{20}{20-r}right) = sum_{r = 0}^{20}a_ra_{20-r},$$
where $a_r = rbinom{20}{r}.$ This is the coefficient of $x^{20}$ in the expansion of $$left(sum_{r=0}^{20}a_rx^rright)^2.$$
answered Feb 28 at 21:21
dezdichado
5,7391928
add a comment |
up vote
2
down vote
up vote
2
down vote
Hint: Write the sum as $$sum_{r=0}^{20}left(rbinom{20}{r}right)cdotleft((20-r)binom{20}{20-r}right) = sum_{r = 0}^{20}a_ra_{20-r},$$
where $a_r = rbinom{20}{r}.$ This is the coefficient of $x^{20}$ in the expansion of $$left(sum_{r=0}^{20}a_rx^rright)^2.$$
answered Feb 28 at 21:21
dezdichado
5,7391928
Hint: Write the sum as $$sum_{r=0}^{20}left(rbinom{20}{r}right)cdotleft((20-r)binom{20}{20-r}right) = sum_{r = 0}^{20}a_ra_{20-r},$$
where $a_r = rbinom{20}{r}.$ This is the coefficient of $x^{20}$ in the expansion of $$left(sum_{r=0}^{20}a_rx^rright)^2.$$
answered Feb 28 at 21:21
dezdichado
5,7391928
edited Feb 28 at 22:20
answered Feb 28 at 21:21
dezdichado
5,7391928
answered Feb 28 at 21:21
dezdichado
5,7391928
answered Feb 28 at 21:21
dezdichado
5,7391928
5,7391928
add a comment |
add a comment |
up vote
0
down vote
$$sum_{r=0}^{20} r(20-r)binom {20}{r} ^2=20sum_{r=0}^{20} rbinom {20}{r} ^2- sum_{r=0}^{20} r^2binom {20}{r} ^2=400sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - 400sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}$$
Hence by Vandermonde's identity we get $$400left(sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}right) =400left(binom {39}{19}- binom {38}{19}right) $$
Now by Pascal's rule $$400left(binom {39}{19}- binom {38}{19}right) =400binom {38}{18}$$
Alternatively you can reach answer by your way too
$$400sum_{color{red}{r=1}}^{color{red}{19}} binom {19}{r}binom {19}{20-r}$$
Hence by using Vandermonde's identity we get $$400sum_{r=1}^{19} binom {19}{r}binom {19}{20-r}=400binom {38}{20}=400binom {38}{18}$$
answered Mar 7 at 16:51
Digamma
5,8241437
add a comment |
up vote
0
down vote
$$sum_{r=0}^{20} r(20-r)binom {20}{r} ^2=20sum_{r=0}^{20} rbinom {20}{r} ^2- sum_{r=0}^{20} r^2binom {20}{r} ^2=400sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - 400sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}$$
Hence by Vandermonde's identity we get $$400left(sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}right) =400left(binom {39}{19}- binom {38}{19}right) $$
Now by Pascal's rule $$400left(binom {39}{19}- binom {38}{19}right) =400binom {38}{18}$$
Alternatively you can reach answer by your way too
$$400sum_{color{red}{r=1}}^{color{red}{19}} binom {19}{r}binom {19}{20-r}$$
Hence by using Vandermonde's identity we get $$400sum_{r=1}^{19} binom {19}{r}binom {19}{20-r}=400binom {38}{20}=400binom {38}{18}$$
answered Mar 7 at 16:51
Digamma
5,8241437
add a comment |
up vote
0
down vote
up vote
0
down vote
$$sum_{r=0}^{20} r(20-r)binom {20}{r} ^2=20sum_{r=0}^{20} rbinom {20}{r} ^2- sum_{r=0}^{20} r^2binom {20}{r} ^2=400sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - 400sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}$$
Hence by Vandermonde's identity we get $$400left(sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}right) =400left(binom {39}{19}- binom {38}{19}right) $$
Now by Pascal's rule $$400left(binom {39}{19}- binom {38}{19}right) =400binom {38}{18}$$
Alternatively you can reach answer by your way too
$$400sum_{color{red}{r=1}}^{color{red}{19}} binom {19}{r}binom {19}{20-r}$$
Hence by using Vandermonde's identity we get $$400sum_{r=1}^{19} binom {19}{r}binom {19}{20-r}=400binom {38}{20}=400binom {38}{18}$$
answered Mar 7 at 16:51
Digamma
5,8241437
$$sum_{r=0}^{20} r(20-r)binom {20}{r} ^2=20sum_{r=0}^{20} rbinom {20}{r} ^2- sum_{r=0}^{20} r^2binom {20}{r} ^2=400sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - 400sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}$$
Hence by Vandermonde's identity we get $$400left(sum_{r=1}^{20} binom {19}{r-1}binom {20}{20-r} - sum_{r=1}^{20} binom {19}{r-1}binom {19}{19-(r-1)}right) =400left(binom {39}{19}- binom {38}{19}right) $$
Now by Pascal's rule $$400left(binom {39}{19}- binom {38}{19}right) =400binom {38}{18}$$
Alternatively you can reach answer by your way too
$$400sum_{color{red}{r=1}}^{color{red}{19}} binom {19}{r}binom {19}{20-r}$$
Hence by using Vandermonde's identity we get $$400sum_{r=1}^{19} binom {19}{r}binom {19}{20-r}=400binom {38}{20}=400binom {38}{18}$$
answered Mar 7 at 16:51
Digamma
5,8241437
edited Mar 7 at 17:13
answered Mar 7 at 16:51
Digamma
5,8241437
answered Mar 7 at 16:51
Digamma
5,8241437
answered Mar 7 at 16:51
Digamma
5,8241437
5,8241437
add a comment |
add a comment |
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Then you replace $binom{19}{r-1}$ by $binom{19}{20-r}$, and think about some combinatorial arguments comparing your summation and $binom{19+19}{r+(20-r)} = binom{38}{20}$... (You have 19 black balls and 19 white balls, in how many ways can you pick 20 balls...?)
– Hw Chu
Feb 28 at 21:14
@HwChu Is it equivalent to the number of non negative integer solutions of $x+y = 20$?
– Abcd
Feb 28 at 21:25
@HwChu I'd like to see the combinatorial solution
– Abcd
Feb 28 at 21:30
1
I wanted to say this is just the Vandermonde's identity. I did not know that identity had a name when I made the comment, so you may just look at the combinarotial proof of Vandermonde's identity.
– Hw Chu
Feb 28 at 21:45