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RiceRiceRice
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Combinatorics Cameron -- Lucas' Theorem Proof
Hi everybody --
Im currently going through Peter Cameron's combinatorics book, and I'm having trouble understanding a step in the proof of Lucas' Theorem, given on page 28 for those of you with the book.
The theorem states for p prime,
m = a0 + a1p + . . . + akpk
n = b0 + b1p + . . . + bkpk
where 0 ≤ ai, bi < p for i = 0, . . ., k -1. Then:
(m choose n ) ≡ [itex]\prod[/itex] (ai choose bi ) (mod p), where the product is taken from i = 0 to i = k.
The proof then states:
It suffices to show that, if m = cp + a and n = dp + b, where 0 ≤ a, b < p, then
(m choose n ) [itex]\equiv[/itex] (c choose d) * (a choose b) (mod p) FOR
a = a0
b = b0
c = a1 + . . . + akpk-1
d = b1 + . . . + bkpk-1
I understand the proof of this statement, but I don't know why proving this is sufficient to proving the original theorem.
Any help would be greatly appreciated! Thanks!
Hi everybody --
Im currently going through Peter Cameron's combinatorics book, and I'm having trouble understanding a step in the proof of Lucas' Theorem, given on page 28 for those of you with the book.
The theorem states for p prime,
m = a0 + a1p + . . . + akpk
n = b0 + b1p + . . . + bkpk
where 0 ≤ ai, bi < p for i = 0, . . ., k -1. Then:
(m choose n ) ≡ [itex]\prod[/itex] (ai choose bi ) (mod p), where the product is taken from i = 0 to i = k.
The proof then states:
It suffices to show that, if m = cp + a and n = dp + b, where 0 ≤ a, b < p, then
(m choose n ) [itex]\equiv[/itex] (c choose d) * (a choose b) (mod p) FOR
a = a0
b = b0
c = a1 + . . . + akpk-1
d = b1 + . . . + bkpk-1
I understand the proof of this statement, but I don't know why proving this is sufficient to proving the original theorem.
Any help would be greatly appreciated! Thanks!