Can Multinomial Coefficients Prove Factorial Inequalities?

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The discussion focuses on proving the inequality |\alpha|! ≤ n^{|\alpha|}α!, where α is a multi-index. The user is attempting to prove a related inequality (a+b)! < 2^{a+b}a!b! using induction, but has faced challenges. Suggestions include using a combinatorial argument and the possibility of applying Stirling's approximation for large n. The user eventually finds a solution and expresses gratitude for the assistance received. The conversation highlights the complexities of factorial inequalities and the utility of combinatorial methods in proofs.
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The actual question is prove that |\alpha|!\le n^{|\alpha|}\alpha! where
\alpha=(\alpha_1,...\alpha_n) is a multi-index (all non-negative) and <br /> |\alpha|=\alpha_1+\cdots +\alpha_n and \alpha!=\alpha_1!\cdots \alpha_n! so I am trying to do it by induction on the number of elements n in \alpha...so I am trying to prove that (a+b)!&lt;2^{a+b}a!b! I have tried to do this by induction on the value of b (the inequality is obvious for b=0 or 1), and other ways, but nothing is working (been trying for close to a week).

Can someone please help? :)

(ps. how do I make it so that after I write in latex it doesn't skip a line like that?)
 
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Are you allowed to use Stirling approximation??

By the way, use [itex ] if you don't want newlines.
 
Anyway, if you're not allowed to use Stirling approximation, just notice that your inequality is equivalent to

\binom{a+b}{a}&lt;2^{a+b}

Now you can use a combinatorial argument.
 
I believe I am allowed to use stirling's approximation, can you suggest a way? (it's only approximate for large n).

Anyway, I will try the other way in the mean time, thanks.
 
Never mind, I got it! You were a huge help, thank you!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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