FMM17: Difference between revisions
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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 < | 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}{ | 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. | 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)/ | 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 | |||
}} | }} | ||
Latest revision as of 18:19, 17 January 2020
Friday Morning Math Problem
A Binomial Problem
$ \binom{16}{r} = \binom{16}{2r+1} $
Find r.
Challenge problem:
$ \binom{120}{r} = \binom{120}{3r+4} $
Find r.
| Solution |
|---|
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First, if we solve it naively, we get
r = 2r + 1 -r = 1 r = -1 We use the fact that $ \binom{n}{m} = \binom{n}{n-m} $ 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 < s <= n such that $ \binom{n}{r} = \binom{n}{s} $ We can prove that $ \binom{n}{r} = \binom{n}{s} $ 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. But to be rigorous we have to cover all our bases. Challenge problem: 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 |
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Friday Morning Math Problems weekly math problems
Spider Socks and Shoes FMM1 Sums of Powers of 2 FMM2 Fifty Coins FMM3 The Zeta Monogram FMM4 The Cthulhus Monogram FMM4B Multiplication Logic FMM5 The Termite and the Cube FMM6 Sharing Dump Trucks FMM7 The Flippant Juror FMM8 Bus Routes FMM9 A Robust Bus System FMM9B Square-Free Sequence FMM10 Inferring Rule from Sequence FMM11 Checkerboard Color Schemes FMM12 One-Handed Chords FMM13 First Ace FMM14 Which Color Cab FMM15 Petersburg Paradox Revisited FMM16 A Binomial Challenge FMM17 A Radical Sum FMM18 Memorable Phone Numbers FMM19 Arrange in Order FMM20 A Pair of Dice Games: FMM21
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