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10 cards from this deck
(a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k(a+b)n=∑k=0n(kn)an−kbk
(nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!}(rn)=r!(n−r)!n!
(1+x)n=∑k=0∞(nk)xk(1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k(1+x)n=∑k=0∞(kn)xk for ∣x∣<1|x|<1∣x∣<1
1−3x+3x2−x31 - 3x + 3x^2 - x^31−3x+3x2−x3
113,983,816\frac{1}{13,983,816}13,983,8161 (choosing 666 from 494949)
The constant term is 64 from (26)(2^6)(26).
Term = (nk)a(n−k)bk\binom{n}{k} a^{(n-k)} b^k(kn)a(n−k)bk
The first term is 81, from (3)4=81(3)^4 = 81(3)4=81.
Coefficient = C(12,6)×(26)=924×64=59136C(12, 6) \times (2^6) = 924 \times 64 = 59136C(12,6)×(26)=924×64=59136
1+12x+60x2+80x3+...1 + 12x + 60x^2 + 80x^3 + ...1+12x+60x2+80x3+...
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