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October 11th, 2017, 06:56 AM   #11
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Quote:
Originally Posted by cjem View Post
This is not true: $(3 \cdot 2)! = 720$ while $(6! - 3!)(3!) = 4284$.
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(5$\displaystyle \cdot$3)! = [15$\displaystyle \cdot$14$\displaystyle \cdot$13$\displaystyle \cdot$12$\displaystyle \cdot$11] [10$\displaystyle \cdot$9$\displaystyle \cdot$8$\displaystyle \cdot$7$\displaystyle \cdot$6] [5$\displaystyle \cdot$4$\displaystyle \cdot$3$\displaystyle \cdot$2$\displaystyle \cdot$1]=[(3p)!/(2p)!] [(2p)!/p!] [p!]
Three (q) terms each of which is divisible by p!. $\displaystyle \therefore$ p!^q is a divisor of (p*q)!

Last edited by zylo; October 11th, 2017 at 07:01 AM.
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October 11th, 2017, 08:30 AM   #12
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(5$\displaystyle \cdot$3)! = [15$\displaystyle \cdot$14$\displaystyle \cdot$13$\displaystyle \cdot$12$\displaystyle \cdot$11] [10$\displaystyle \cdot$9$\displaystyle \cdot$8$\displaystyle \cdot$7$\displaystyle \cdot$6] [5$\displaystyle \cdot$4$\displaystyle \cdot$3$\displaystyle \cdot$2$\displaystyle \cdot$1]=[(3p)!/(2p)!] [(2p)!/p!] [p!]
Three (q) terms each of which is divisible by p!. $\displaystyle \therefore$ p!^q is a divisor of (p*q)!
The example is clear, but the generalization isn't. For example,

(5$\displaystyle \cdot$3)!/5! is indeed an integer, but you don't know that

(5$\displaystyle \cdot$3)!/[5!(5$\displaystyle \cdot$2)!] is.
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