October 11th, 2017, 06:56 AM  #11  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
(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.  
October 11th, 2017, 08:30 AM  #12  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
(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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