Combinatorial proof identity

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Give a combinatorial proof of the following identity: $$binom3n3 =3binomn3 +6nbinomn2 +n^3.$$

I've been working on this proof for hours, however I'm not able to show LHS = RHS-
I completely understand binomial theorem and few combinatorial proofs but not able to succeed this one.
Help would be appreciated.

edited 44 mins ago

Tianlalu

2,001629

asked 59 mins ago

m.saza

202

add a commentÂ |Â

up vote
2
down vote

favorite

Give a combinatorial proof of the following identity: $$binom3n3 =3binomn3 +6nbinomn2 +n^3.$$

edited 44 mins ago

Tianlalu

2,001629

asked 59 mins ago

m.saza

202

add a commentÂ |Â

up vote
2
down vote

favorite

Give a combinatorial proof of the following identity: $$binom3n3 =3binomn3 +6nbinomn2 +n^3.$$

edited 44 mins ago

Tianlalu

2,001629

asked 59 mins ago

m.saza

202

Give a combinatorial proof of the following identity: $$binom3n3 =3binomn3 +6nbinomn2 +n^3.$$

combinatorics

edited 44 mins ago

Tianlalu

2,001629

asked 59 mins ago

m.saza

202

edited 44 mins ago

Tianlalu

2,001629

asked 59 mins ago

m.saza

202

edited 44 mins ago

Tianlalu

2,001629

edited 44 mins ago

Tianlalu

2,001629

edited 44 mins ago

Tianlalu

2,001629

asked 59 mins ago

m.saza

202

asked 59 mins ago

m.saza

202

asked 59 mins ago

m.saza

202

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
4
down vote

Arrange $3n$ balls into 3 rows and each row contains $n$ balls.

There are $binomn3$ ways to select $3$ balls from them. We can group the ways into $3$ categories:

Select one ball from each row. There are $n$ choices for each row, this contribute $n^3$ ways of pick the balls.

Select two balls from one row and one ball from another row. There are $3 times 2$ ways to select the rows. Since there are $binomn2$ ways to select two balls from a row and $n$ ways to select one balls from a row, this contribute $6 binomn2 n$ ways to pick the balls.

Select three balls from a single row. There are $3$ ways to select the row and $binomn3$ ways to select three balls from that particular row. This contributes $3binomn3$ ways.

These $3$ categories of choicing doesn't overlap and exhaust all possible ways to select three balls. As a result,

$$binom3n3 = n^3 + 6binomn2 n + 3binomn3$$

answered 41 mins ago

achille hui

92.2k5127248

Great answer +1!
â€“Â Rushabh Mehta
36 mins ago

add a commentÂ |Â

up vote
2
down vote

$$3nchoose3=3nchoose3+6nnchoose2+n^3$$$$frac12ncdot(3n-1)cdot(3n-2)=frac12ncdot(n-1)cdot(n-2)+3n^2cdot(n-1)+n^3$$Can you take it from here?

edited 41 mins ago

answered 51 mins ago

Rushabh Mehta

3,760530

Why do people post algebraic solutions when the OP is asking specifically (in boldface) for a combinatorial proof? Not gonna vote down algebraic solutions though. Just saying. And +1 only to Achille Hui.
â€“Â Ken Draco
19 mins ago

1

@KenDraco I might have misread the question at first. Oops!
â€“Â Rushabh Mehta
17 mins ago

I got it. I myself am prone to such things. I just mean all the answers below -:)
â€“Â Ken Draco
8 mins ago

add a commentÂ |Â

up vote
0
down vote

$2!=2, 3!=6$, so:

$$binom3n3=frac n2(3n-1)(3n-2)$$
$$3binomn3=frac n2(n-1)(n-2)$$
$$6nbinomn2=3n^2(n-1)$$

I used that $$fracx!(x-a)!=x^underlinex-a=prod_n=0^a-1(x-n)$$

See falling factorials

answered 48 mins ago

Rhys Hughes

4,3741327

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
41 mins ago

add a commentÂ |Â

up vote
0
down vote

Brute force:

$27n^3 - 27n^2+ 6n = 27n^3 - 27n^2 + 6n$

$27n^3 - 9n^2 - 18n^2 + 6n = 3n^3 - 3n^2 - 6n^2 + 6n + 18n^3 - 18n^2 + 6n^3$

$(9n^2 - 3n)(3n - 2) = (3n^2 - 3n)(n - 2) + 18n^2(n-1) + 6n^3$

$3n(3n - 1)(3n - 2) = 3n(n - 1)(n - 2) + 3(6n)n(n-1) + 6n^3$

$3n(3n - 1)(3n - 2)/6 = 3n(n - 1)(n - 2)/6 + (6n)n(n-1)/2 + n^3$

$binom3n3 =3binomn3 +6nbinomn2 +n^3$

edited 43 mins ago

answered 50 mins ago

R zu

2379

1

A proof that thinks about combination (as in choosing n objects from m objects) is better because it trains intuition. But for simple problems, algebra and brute force is sufficient.
â€“Â R zu
48 mins ago

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
40 mins ago

add a commentÂ |Â

up vote
0
down vote

Notice that the LHS is:

$$frac3n times (3n-1) times (3n-2)3! = fracn times (3n-1) times (3n-2)2$$
$$to frac9n^3-9n^2+2n2$$

For the RHS:

$$3 times fracntimes(n-1)times(n-2)3! + 6n times fracntimes(n-1)2! + frac2n^32$$

$$Longrightarrow fracntimes(n-1)times(n-2)2 + frac6n^2 times(n-1)2 + frac2n^32$$

$$to frac9n^3-9n^2+2n2$$

Hence, RHS=LHS.

edited 21 mins ago

answered 30 mins ago

Maged Saeed

419215

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
4
down vote

Arrange $3n$ balls into 3 rows and each row contains $n$ balls.

There are $binomn3$ ways to select $3$ balls from them. We can group the ways into $3$ categories:

Select one ball from each row. There are $n$ choices for each row, this contribute $n^3$ ways of pick the balls.

Select two balls from one row and one ball from another row. There are $3 times 2$ ways to select the rows. Since there are $binomn2$ ways to select two balls from a row and $n$ ways to select one balls from a row, this contribute $6 binomn2 n$ ways to pick the balls.

Select three balls from a single row. There are $3$ ways to select the row and $binomn3$ ways to select three balls from that particular row. This contributes $3binomn3$ ways.

These $3$ categories of choicing doesn't overlap and exhaust all possible ways to select three balls. As a result,

$$binom3n3 = n^3 + 6binomn2 n + 3binomn3$$

answered 41 mins ago

achille hui

92.2k5127248

Great answer +1!
â€“Â Rushabh Mehta
36 mins ago

add a commentÂ |Â

up vote
4
down vote

Arrange $3n$ balls into 3 rows and each row contains $n$ balls.

There are $binomn3$ ways to select $3$ balls from them. We can group the ways into $3$ categories:

Select one ball from each row. There are $n$ choices for each row, this contribute $n^3$ ways of pick the balls.

Select two balls from one row and one ball from another row. There are $3 times 2$ ways to select the rows. Since there are $binomn2$ ways to select two balls from a row and $n$ ways to select one balls from a row, this contribute $6 binomn2 n$ ways to pick the balls.

Select three balls from a single row. There are $3$ ways to select the row and $binomn3$ ways to select three balls from that particular row. This contributes $3binomn3$ ways.

These $3$ categories of choicing doesn't overlap and exhaust all possible ways to select three balls. As a result,

$$binom3n3 = n^3 + 6binomn2 n + 3binomn3$$

answered 41 mins ago

achille hui

92.2k5127248

Great answer +1!
â€“Â Rushabh Mehta
36 mins ago

add a commentÂ |Â

up vote
4
down vote

Arrange $3n$ balls into 3 rows and each row contains $n$ balls.

There are $binomn3$ ways to select $3$ balls from them. We can group the ways into $3$ categories:

Select one ball from each row. There are $n$ choices for each row, this contribute $n^3$ ways of pick the balls.

Select two balls from one row and one ball from another row. There are $3 times 2$ ways to select the rows. Since there are $binomn2$ ways to select two balls from a row and $n$ ways to select one balls from a row, this contribute $6 binomn2 n$ ways to pick the balls.

Select three balls from a single row. There are $3$ ways to select the row and $binomn3$ ways to select three balls from that particular row. This contributes $3binomn3$ ways.

These $3$ categories of choicing doesn't overlap and exhaust all possible ways to select three balls. As a result,

$$binom3n3 = n^3 + 6binomn2 n + 3binomn3$$

answered 41 mins ago

achille hui

92.2k5127248

Arrange $3n$ balls into 3 rows and each row contains $n$ balls.

There are $binomn3$ ways to select $3$ balls from them. We can group the ways into $3$ categories:

Select one ball from each row. There are $n$ choices for each row, this contribute $n^3$ ways of pick the balls.

Select two balls from one row and one ball from another row. There are $3 times 2$ ways to select the rows. Since there are $binomn2$ ways to select two balls from a row and $n$ ways to select one balls from a row, this contribute $6 binomn2 n$ ways to pick the balls.

Select three balls from a single row. There are $3$ ways to select the row and $binomn3$ ways to select three balls from that particular row. This contributes $3binomn3$ ways.

These $3$ categories of choicing doesn't overlap and exhaust all possible ways to select three balls. As a result,

$$binom3n3 = n^3 + 6binomn2 n + 3binomn3$$

answered 41 mins ago

achille hui

92.2k5127248

answered 41 mins ago

achille hui

92.2k5127248

answered 41 mins ago

achille hui

92.2k5127248

answered 41 mins ago

achille hui

92.2k5127248

Great answer +1!
â€“Â Rushabh Mehta
36 mins ago

add a commentÂ |Â

Great answer +1!
â€“Â Rushabh Mehta
36 mins ago

Great answer +1!
â€“Â Rushabh Mehta
36 mins ago

add a commentÂ |Â

up vote
2
down vote

$$3nchoose3=3nchoose3+6nnchoose2+n^3$$$$frac12ncdot(3n-1)cdot(3n-2)=frac12ncdot(n-1)cdot(n-2)+3n^2cdot(n-1)+n^3$$Can you take it from here?

edited 41 mins ago

answered 51 mins ago

Rushabh Mehta

3,760530

Why do people post algebraic solutions when the OP is asking specifically (in boldface) for a combinatorial proof? Not gonna vote down algebraic solutions though. Just saying. And +1 only to Achille Hui.
â€“Â Ken Draco
19 mins ago

1

@KenDraco I might have misread the question at first. Oops!
â€“Â Rushabh Mehta
17 mins ago

I got it. I myself am prone to such things. I just mean all the answers below -:)
â€“Â Ken Draco
8 mins ago

add a commentÂ |Â

up vote
2
down vote

$$3nchoose3=3nchoose3+6nnchoose2+n^3$$$$frac12ncdot(3n-1)cdot(3n-2)=frac12ncdot(n-1)cdot(n-2)+3n^2cdot(n-1)+n^3$$Can you take it from here?

edited 41 mins ago

answered 51 mins ago

Rushabh Mehta

3,760530

Why do people post algebraic solutions when the OP is asking specifically (in boldface) for a combinatorial proof? Not gonna vote down algebraic solutions though. Just saying. And +1 only to Achille Hui.
â€“Â Ken Draco
19 mins ago

1

@KenDraco I might have misread the question at first. Oops!
â€“Â Rushabh Mehta
17 mins ago

I got it. I myself am prone to such things. I just mean all the answers below -:)
â€“Â Ken Draco
8 mins ago

add a commentÂ |Â

up vote
2
down vote

$$3nchoose3=3nchoose3+6nnchoose2+n^3$$$$frac12ncdot(3n-1)cdot(3n-2)=frac12ncdot(n-1)cdot(n-2)+3n^2cdot(n-1)+n^3$$Can you take it from here?

edited 41 mins ago

answered 51 mins ago

Rushabh Mehta

3,760530

$$3nchoose3=3nchoose3+6nnchoose2+n^3$$$$frac12ncdot(3n-1)cdot(3n-2)=frac12ncdot(n-1)cdot(n-2)+3n^2cdot(n-1)+n^3$$Can you take it from here?

edited 41 mins ago

answered 51 mins ago

Rushabh Mehta

3,760530

edited 41 mins ago

answered 51 mins ago

Rushabh Mehta

3,760530

answered 51 mins ago

Rushabh Mehta

3,760530

answered 51 mins ago

Rushabh Mehta

3,760530

Why do people post algebraic solutions when the OP is asking specifically (in boldface) for a combinatorial proof? Not gonna vote down algebraic solutions though. Just saying. And +1 only to Achille Hui.
â€“Â Ken Draco
19 mins ago

1

@KenDraco I might have misread the question at first. Oops!
â€“Â Rushabh Mehta
17 mins ago

I got it. I myself am prone to such things. I just mean all the answers below -:)
â€“Â Ken Draco
8 mins ago

add a commentÂ |Â

Why do people post algebraic solutions when the OP is asking specifically (in boldface) for a combinatorial proof? Not gonna vote down algebraic solutions though. Just saying. And +1 only to Achille Hui.
â€“Â Ken Draco
19 mins ago

1

@KenDraco I might have misread the question at first. Oops!
â€“Â Rushabh Mehta
17 mins ago

I got it. I myself am prone to such things. I just mean all the answers below -:)
â€“Â Ken Draco
8 mins ago

Why do people post algebraic solutions when the OP is asking specifically (in boldface) for a combinatorial proof? Not gonna vote down algebraic solutions though. Just saying. And +1 only to Achille Hui.
â€“Â Ken Draco
19 mins ago

@KenDraco I might have misread the question at first. Oops!
â€“Â Rushabh Mehta
17 mins ago

I got it. I myself am prone to such things. I just mean all the answers below -:)
â€“Â Ken Draco
8 mins ago

add a commentÂ |Â

up vote
0
down vote

$2!=2, 3!=6$, so:

$$binom3n3=frac n2(3n-1)(3n-2)$$
$$3binomn3=frac n2(n-1)(n-2)$$
$$6nbinomn2=3n^2(n-1)$$

I used that $$fracx!(x-a)!=x^underlinex-a=prod_n=0^a-1(x-n)$$

See falling factorials

answered 48 mins ago

Rhys Hughes

4,3741327

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
41 mins ago

add a commentÂ |Â

up vote
0
down vote

$2!=2, 3!=6$, so:

$$binom3n3=frac n2(3n-1)(3n-2)$$
$$3binomn3=frac n2(n-1)(n-2)$$
$$6nbinomn2=3n^2(n-1)$$

I used that $$fracx!(x-a)!=x^underlinex-a=prod_n=0^a-1(x-n)$$

See falling factorials

answered 48 mins ago

Rhys Hughes

4,3741327

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
41 mins ago

add a commentÂ |Â

up vote
0
down vote

$2!=2, 3!=6$, so:

$$binom3n3=frac n2(3n-1)(3n-2)$$
$$3binomn3=frac n2(n-1)(n-2)$$
$$6nbinomn2=3n^2(n-1)$$

I used that $$fracx!(x-a)!=x^underlinex-a=prod_n=0^a-1(x-n)$$

See falling factorials

answered 48 mins ago

Rhys Hughes

4,3741327

$2!=2, 3!=6$, so:

$$binom3n3=frac n2(3n-1)(3n-2)$$
$$3binomn3=frac n2(n-1)(n-2)$$
$$6nbinomn2=3n^2(n-1)$$

I used that $$fracx!(x-a)!=x^underlinex-a=prod_n=0^a-1(x-n)$$

See falling factorials

answered 48 mins ago

Rhys Hughes

4,3741327

answered 48 mins ago

Rhys Hughes

4,3741327

answered 48 mins ago

Rhys Hughes

4,3741327

answered 48 mins ago

Rhys Hughes

4,3741327

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
41 mins ago

add a commentÂ |Â

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
41 mins ago

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
41 mins ago

add a commentÂ |Â

up vote
0
down vote

Brute force:

$27n^3 - 27n^2+ 6n = 27n^3 - 27n^2 + 6n$

$27n^3 - 9n^2 - 18n^2 + 6n = 3n^3 - 3n^2 - 6n^2 + 6n + 18n^3 - 18n^2 + 6n^3$

$(9n^2 - 3n)(3n - 2) = (3n^2 - 3n)(n - 2) + 18n^2(n-1) + 6n^3$

$3n(3n - 1)(3n - 2) = 3n(n - 1)(n - 2) + 3(6n)n(n-1) + 6n^3$

$3n(3n - 1)(3n - 2)/6 = 3n(n - 1)(n - 2)/6 + (6n)n(n-1)/2 + n^3$

$binom3n3 =3binomn3 +6nbinomn2 +n^3$

edited 43 mins ago

answered 50 mins ago

R zu

2379

1

A proof that thinks about combination (as in choosing n objects from m objects) is better because it trains intuition. But for simple problems, algebra and brute force is sufficient.
â€“Â R zu
48 mins ago

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
40 mins ago

add a commentÂ |Â

up vote
0
down vote

Brute force:

$27n^3 - 27n^2+ 6n = 27n^3 - 27n^2 + 6n$

$27n^3 - 9n^2 - 18n^2 + 6n = 3n^3 - 3n^2 - 6n^2 + 6n + 18n^3 - 18n^2 + 6n^3$

$(9n^2 - 3n)(3n - 2) = (3n^2 - 3n)(n - 2) + 18n^2(n-1) + 6n^3$

$3n(3n - 1)(3n - 2) = 3n(n - 1)(n - 2) + 3(6n)n(n-1) + 6n^3$

$3n(3n - 1)(3n - 2)/6 = 3n(n - 1)(n - 2)/6 + (6n)n(n-1)/2 + n^3$

$binom3n3 =3binomn3 +6nbinomn2 +n^3$

edited 43 mins ago

answered 50 mins ago

R zu

2379

1

A proof that thinks about combination (as in choosing n objects from m objects) is better because it trains intuition. But for simple problems, algebra and brute force is sufficient.
â€“Â R zu
48 mins ago

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
40 mins ago

add a commentÂ |Â

up vote
0
down vote

Brute force:

$27n^3 - 27n^2+ 6n = 27n^3 - 27n^2 + 6n$

$27n^3 - 9n^2 - 18n^2 + 6n = 3n^3 - 3n^2 - 6n^2 + 6n + 18n^3 - 18n^2 + 6n^3$

$(9n^2 - 3n)(3n - 2) = (3n^2 - 3n)(n - 2) + 18n^2(n-1) + 6n^3$

$3n(3n - 1)(3n - 2) = 3n(n - 1)(n - 2) + 3(6n)n(n-1) + 6n^3$

$3n(3n - 1)(3n - 2)/6 = 3n(n - 1)(n - 2)/6 + (6n)n(n-1)/2 + n^3$

$binom3n3 =3binomn3 +6nbinomn2 +n^3$

edited 43 mins ago

answered 50 mins ago

R zu

2379

Brute force:

$27n^3 - 27n^2+ 6n = 27n^3 - 27n^2 + 6n$

$27n^3 - 9n^2 - 18n^2 + 6n = 3n^3 - 3n^2 - 6n^2 + 6n + 18n^3 - 18n^2 + 6n^3$

$(9n^2 - 3n)(3n - 2) = (3n^2 - 3n)(n - 2) + 18n^2(n-1) + 6n^3$

$3n(3n - 1)(3n - 2) = 3n(n - 1)(n - 2) + 3(6n)n(n-1) + 6n^3$

$3n(3n - 1)(3n - 2)/6 = 3n(n - 1)(n - 2)/6 + (6n)n(n-1)/2 + n^3$

$binom3n3 =3binomn3 +6nbinomn2 +n^3$

edited 43 mins ago

answered 50 mins ago

R zu

2379

edited 43 mins ago

answered 50 mins ago

R zu

2379

answered 50 mins ago

R zu

2379

answered 50 mins ago

R zu

2379

1

A proof that thinks about combination (as in choosing n objects from m objects) is better because it trains intuition. But for simple problems, algebra and brute force is sufficient.
â€“Â R zu
48 mins ago

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
40 mins ago

add a commentÂ |Â

1

A proof that thinks about combination (as in choosing n objects from m objects) is better because it trains intuition. But for simple problems, algebra and brute force is sufficient.
â€“Â R zu
48 mins ago

2

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
40 mins ago

A proof that thinks about combination (as in choosing n objects from m objects) is better because it trains intuition. But for simple problems, algebra and brute force is sufficient.
â€“Â R zu
48 mins ago

This would be an algebraic proof, not a combinatorial proof.
â€“Â Mark S.
40 mins ago

add a commentÂ |Â

up vote
0
down vote

Notice that the LHS is:

$$frac3n times (3n-1) times (3n-2)3! = fracn times (3n-1) times (3n-2)2$$
$$to frac9n^3-9n^2+2n2$$

For the RHS:

$$3 times fracntimes(n-1)times(n-2)3! + 6n times fracntimes(n-1)2! + frac2n^32$$

$$Longrightarrow fracntimes(n-1)times(n-2)2 + frac6n^2 times(n-1)2 + frac2n^32$$

$$to frac9n^3-9n^2+2n2$$

Hence, RHS=LHS.

edited 21 mins ago

answered 30 mins ago

Maged Saeed

419215

add a commentÂ |Â

up vote
0
down vote

Notice that the LHS is:

$$frac3n times (3n-1) times (3n-2)3! = fracn times (3n-1) times (3n-2)2$$
$$to frac9n^3-9n^2+2n2$$

For the RHS:

$$3 times fracntimes(n-1)times(n-2)3! + 6n times fracntimes(n-1)2! + frac2n^32$$

$$Longrightarrow fracntimes(n-1)times(n-2)2 + frac6n^2 times(n-1)2 + frac2n^32$$

$$to frac9n^3-9n^2+2n2$$

Hence, RHS=LHS.

edited 21 mins ago

answered 30 mins ago

Maged Saeed

419215

add a commentÂ |Â

up vote
0
down vote

Notice that the LHS is:

$$frac3n times (3n-1) times (3n-2)3! = fracn times (3n-1) times (3n-2)2$$
$$to frac9n^3-9n^2+2n2$$

For the RHS:

$$3 times fracntimes(n-1)times(n-2)3! + 6n times fracntimes(n-1)2! + frac2n^32$$

$$Longrightarrow fracntimes(n-1)times(n-2)2 + frac6n^2 times(n-1)2 + frac2n^32$$

$$to frac9n^3-9n^2+2n2$$

Hence, RHS=LHS.

edited 21 mins ago

answered 30 mins ago

Maged Saeed

419215

Notice that the LHS is:

$$frac3n times (3n-1) times (3n-2)3! = fracn times (3n-1) times (3n-2)2$$
$$to frac9n^3-9n^2+2n2$$

For the RHS:

$$3 times fracntimes(n-1)times(n-2)3! + 6n times fracntimes(n-1)2! + frac2n^32$$

$$Longrightarrow fracntimes(n-1)times(n-2)2 + frac6n^2 times(n-1)2 + frac2n^32$$

$$to frac9n^3-9n^2+2n2$$

Hence, RHS=LHS.

edited 21 mins ago

answered 30 mins ago

Maged Saeed

419215

edited 21 mins ago

answered 30 mins ago

Maged Saeed

419215

answered 30 mins ago

Maged Saeed

419215

answered 30 mins ago

Maged Saeed

419215

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