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Divisibility factorial

Prove that \(\cfrac { (3n)! }{ { (3!) }^{ n } }\) is integer for all \(n\ge 0\).

This is the solution I tried

\({(3!)}^{ n }={(6) }^{ n }\).We have to prove \({(6) }^{ n }|(3n)!\).We have product of 3 consecutive numbers is divisible by \(6\).Now there are \(n\) pairs of \(3\) consecutive numbers.Therefore \({ 6 }^{ n }y=(3n)!\) for some \(y\)..Therefore \(\cfrac { (3n)! }{ { (3!) }^{ n } }\) is integer for all \(n\ge 0\).

Is it correct?

Note by Shivamani Patil
2 years, 5 months ago

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The given expression can be written as \[\dfrac{(3n)!}{6^{n}}\] which can be written as \[\dfrac{(3n)!}{3^{n}\times2^{n}}.\]Now,we have that the maximum power of \(3\) that divides \((3n!)\)\[=\lfloor{\dfrac{3n}{3}}\rfloor+\lfloor{\dfrac{3n}{9}}\rfloor+...\]Similarly the maximum power of \(2\) that divides \(3n!\)=\[\lfloor{\dfrac{3n}{2}}\rfloor+\lfloor{\dfrac{3n}{4}}\rfloor+.....\]Now,for\(n\geq3\) the expressions above are \(>3^{n}\) and \(2^{n}\) respectively.Thu for all\(n\geq3\) the given expression is an integer.Checking cases for \(n=0,1,2\) reveals that the given expression is an integer for all three.Hence proved. Adarsh Kumar · 2 years, 5 months ago

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Using some combinatorics:

(3n)! / (3!)^n = (3n)! / (3!)(3!) ... (3!)(3!) n factors of (3!). One can think of this as the number of identical permutations with 3n letters and n different letters with 3 as frequency in the letter. It remains to prove that the number of identical permutations is an integer. John Ashley Capellan · 2 years, 5 months ago

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@John Ashley Capellan Your solution is great.But can you give elementary number theory proof. Shivamani Patil · 2 years, 5 months ago

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I believe I have a brilliantly simple solution to your question - I will use induction. Let f(n) = \(\frac{(3n)!}{(3!)^{n} }\). Base test: for n=1, f(1) = 1 and f(2) = \(\frac{6!}{3! * 3!}\) = \(\frac{4 * 5 * 6}{6}\) = 20. Now for the inductive step: \(\frac{(3n+3)!}{(3!)^{n} * 3!}\), and by breaking up the fraction we see that f(n+1) is a product of \(\frac{(3n)!}{(3!)^{n} }\) and \(\frac{(3n+1)(3n+2)(3n+3)}{3!}\). We already know that the former is already an integer by our inductive hypothesis, so we need to prove that the latter is also an integer. This is simple as 3 or more consecutive integers are divisible by 3 and 2 (In fact, n consecutive integers are always divisible by n!). Hence, our proof is complete. I hope that helps :D Curtis Clement · 2 years, 4 months ago

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Note that \(\dfrac{(3n)!}{(3!)^n}\) represents the number of ways to arrange \(3n\) objects with \(n\) triplets of object being identical, in a row. For instance, at \(n = 2\) , you are actually arranging 6 objects with 2 triplets, such as arranging AAABBB in a row, yielding \(\dfrac{6!}{3!3!} = \dfrac{(2(3))!}{(3!)^2}\)

Note that the number of objects = number of objects limited by the requirement of the permutation. (That is from the previous example, if I have \(6\) objects , and there must be \(2\) sets of \(3\) objects, it is physically possible.) Hence, it is actually possible to arrange \(3n\) objects with \(n\) triplets of object being identical in a row, thus causing the number of ways to be an integer. Henceforth, \(\dfrac{(3n)!}{(3!)^n}\) that represents the number of ways to arrange \(3n\) objects with \(n\) triplets of object being identical, in a row, will thus be an integer.

P.S : If however, the number of objects \(<\) number of objects limited by the requirement of the permutation, the scenario will not physically happens. Hence, the expression that evaluates the number of ways might not be an integer. However, since we are not interested in it, we can hence ignore it. Tan Kiat · 2 years, 5 months ago

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(3!)^n=3^n×2^n.now 3n> 3 (n-1)> 3 (n-2)......> 3 so 3n×3 (n-1)×3 (n-2).........×3=3^n×n! contains in (3n)!.again 3n> 2n> 2 (n-1).......> 2 so 2n×2 (n-1)......×2=2^n×n! contains in (3n)!.so (3n)! is is divisible by (3!)^n Subhrajyoti Sinha · 2 years, 5 months ago

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@Subhrajyoti Sinha Try formatting your maths so it's easier to read - I know it's tricky so I'll give you a few pointers. For (3)!^n place curly brackets around the n (i.e. {n} ), then place normal brackets around the whole expression, before putting a '\' before you 1st and last bracket - \(...\). This should give \((3!)^{n}\). Also, (3n)! = 3n(3n-1)(3n-2)...2*1 as it is inside the brackets, rather than outside. Curtis Clement · 2 years, 4 months ago

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There are \( n(3n-1)! \) pairs of 3 consecutive numbers.

How did get this? Since there are \( 3n \) numbers being multiplied, there should be \( \frac{3n}{3} = n \) pairs of consecutive number. For example, \( 6! =(6*5*4)(3*2*1) \) can be partitioned into 2 pairs. Siddhartha Srivastava · 2 years, 5 months ago

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@Siddhartha Srivastava But we have to take \((3n)!\) numbers in pairs of 3 .Therefore dividing by 3 we get \(n(3n-1)!\) pairs. Shivamani Patil · 2 years, 5 months ago

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@Shivamani Patil \( (3n)! \) is the product of \( 3n \) numbers,i.e \( 1*2*...*(3n-1)*(3n) \). Not \( (3n)! \) numbers. Also, you can check smaller cases. The highest power of \( 6 \) dividing \( (3*2)!\) is \( 6^2 \), not \( 6^{2(5!)} = 6^{240} \). Siddhartha Srivastava · 2 years, 5 months ago

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@Siddhartha Srivastava Ohh yes I missed it in hurry .I have updated solution now is it correct? Shivamani Patil · 2 years, 5 months ago

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@Shivamani Patil Yes, now it is, though you could also mention the fact that the product of 3 consecutive integers is also divisible by 6. Siddhartha Srivastava · 2 years, 5 months ago

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@Siddhartha Srivastava I have mentioned that na. Shivamani Patil · 2 years, 5 months ago

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@Shivamani Patil Sorry. Didn't read. Siddhartha Srivastava · 2 years, 5 months ago

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@Siddhartha Srivastava That's ok. u study in which class? Shivamani Patil · 2 years, 5 months ago

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@Shivamani Patil 11th. Siddhartha Srivastava · 2 years, 5 months ago

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@Siddhartha Srivastava Ohhh Shivamani Patil · 2 years, 5 months ago

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