FMM17 - Revision history

Admin at 18:19, 17 January 2020

2020-01-17T18:19:26Z

← Older revision		Revision as of 18:19, 17 January 2020
Line 39:		Line 39:
	r = 5		r = 5

	What more do we need to do? We can use this identity to transform r = 2r + 1, but there may also exist other integers 0 <= r < l <= n such that <math>\binom{n}{r} = \binom{n}{l}</math>		What more do we need to do? We can use this identity to transform r = 2r + 1, but there may also exist other integers 0 <= r < s <= n such that <math>\binom{n}{r} = \binom{n}{s}</math>

	We can prove that <math>\binom{n}{r} = \binom{n}{l}</math> only holds if r = l or r = n - l		We can prove that <math>\binom{n}{r} = \binom{n}{s}</math> only holds if r = s or r = n - s

	This is something we may implicitly assume, due to our own understanding of the nature of the binomial numbers and Pascal's Triangle. ~~We just~~ have to cover all our bases.		This is something we may implicitly assume, due to our own understanding of the nature of the binomial numbers and Pascal's Triangle. But to be rigorous we have to cover all our bases.

	Challenge problem:		Challenge problem:

	r = (120 - 4)/~~(1 + 3)~~ = 29		Solving naively we get

			r = 3r + 4

			-4r = 4

			r = -1

			To transform, use the same identity as above

			3r + 4 = r

			3r + 4 = 120 - r

			4r = 120 - 4

			r = (120 - 4)/4 = 29

	}}		}}

Admin at 18:14, 17 January 2020

2020-01-17T18:14:58Z

← Older revision		Revision as of 18:14, 17 January 2020
Line 4:		Line 4:

	<math>		<math>
	\binom{16}{4} = \binom{16}{2r+1}		\binom{16}{r} = \binom{16}{2r+1}
	</math>		</math>

Line 17:		Line 17:
	Find r.		Find r.

	\|~~answer~~=		\|solution=

	First, if we solve it naively, we get		First, if we solve it naively, we get

Admin at 10:32, 4 December 2019

2019-12-04T10:32:50Z

← Older revision		Revision as of 10:32, 4 December 2019
Line 5:		Line 5:
	<math>		<math>
	\binom{16}{4} = \binom{16}{2r+1}		\binom{16}{4} = \binom{16}{2r+1}
			</math>

			Find r.

			Challenge problem:

			<math>
			\binom{120}{r} = \binom{120}{3r+4}
	</math>		</math>

Line 36:		Line 44:

	This is something we may implicitly assume, due to our own understanding of the nature of the binomial numbers and Pascal's Triangle. We just have to cover all our bases.		This is something we may implicitly assume, due to our own understanding of the nature of the binomial numbers and Pascal's Triangle. We just have to cover all our bases.

			Challenge problem:

			r = (120 - 4)/(1 + 3) = 29

	}}		}}

Admin: Created page with "{{FMM |title=A Binomial Problem |problem= $\binom{16}{4} = \binom{16}{2r+1}$ Find r. |answer= First, if we solve it naively, we get r = 2r + 1 -r = 1 r = -..."

2019-12-04T10:29:59Z

Created page with "{{FMM |title=A Binomial Problem |problem= <math> \binom{16}{4} = \binom{16}{2r+1} </math> Find r. |answer= First, if we solve it naively, we get r = 2r + 1 -r = 1 r = -..."

New page

{{FMM
|title=A Binomial Problem
|problem=

<math>
\binom{16}{4} = \binom{16}{2r+1}
</math>

Find r.

|answer=

First, if we solve it naively, we get

r = 2r + 1

-r = 1

r = -1

We use the fact that <math>\binom{n}{m} = \binom{n}{n-m}</math>

2r + 1 =16 - (2r+1)

r = 16 - (2r+1)

r = 16 - 2r - 1

3r = 15

r = 5

What more do we need to do? We can use this identity to transform r = 2r + 1, but there may also exist other integers 0 <= r < l <= n such that <math>\binom{n}{r} = \binom{n}{l}</math>

We can prove that <math>\binom{n}{r} = \binom{n}{l}</math> only holds if r = l or r = n - l

This is something we may implicitly assume, due to our own understanding of the nature of the binomial numbers and Pascal's Triangle. We just have to cover all our bases.

}}

FMM17 - Revision history

Admin at 18:19, 17 January 2020

Admin at 18:14, 17 January 2020

Admin at 10:32, 4 December 2019

Admin: Created page with "{{FMM |title=A Binomial Problem |problem= \binom{16}{4} = \binom{16}{2r+1} Find r. |answer= First, if we solve it naively, we get r = 2r + 1 -r = 1 r = -..."

Admin: Created page with "{{FMM |title=A Binomial Problem |problem= $\binom{16}{4} = \binom{16}{2r+1}$ Find r. |answer= First, if we solve it naively, we get r = 2r + 1 -r = 1 r = -..."