Vtyjkyuk

A friend of mine asked me to prove that $$frac200!(10!)^20$$ is an integer

I used a basic example in which I assumed that there are $200$ objects places in $20$ boxes (which means that effectively there are $10$ objects in one box). One more condition that I adopted was that the boxes are distinguishable but the items within each box are not. Now the number of permutations possible for such an arrangement are :
$$ frac200!underbrace10! cdot 10! cdot 10!cdots 10!_text$20$ times$$
$$Rightarrow frac200!(10!) ^20$$

Since these are just ways of arranging, we can be pretty sure that this number is an integer.

Then he made the problem more complex by adding a $19!$ in the denominator, thus making the problem:
Is $$frac200!(10!)^20 cdot 19!$$ an integer or not?

The $19!$ in the denominator seemed to be pretty odd and hence I couldn't find any intuitive way to determine the thing. Can anybody please help me with the problem?

You assumed the boxes were distinguishable, leading to $frac200!(10!)^20$, ways to fill the boxes. If you make them indistinguishable, you merge the $20!$ ways of reordering the boxes into one, so that previous answer overcounts each way of filling indistinguishable boxes by a factor of $20!$. Therefore you are left with $frac200!(10!)^20/20!$ ways to fill 20 indistinguishable boxes, which then must be an integer. After multiplying by $20$ it is of course still an integer.

We know that $dfrac(mn)!n!(m!)^n$ is an integer for $m,n in Bbb N$ $^(*)$ . Let $n = 20$ and $m = 10$, then $dfrac(200)!20!(10!)^20$ is an integer.

Multiply by $20$, $dfrac(200)!19!(10!)^20$ is an integer.

$(*)$ : Prove that $(mn)!$ is divisible by $(n!)cdot(m!)^n$

Using induction, this answer says that
$$
frac(mn)!(m!)^nn!=prod_k=1^nbinommk-1m-1
$$
Plug in $m=10$ and $n=20$ to get
$$
frac200!10!^20,20!=prod_k=1^20binom10k-19
$$
Multiply by $20$ to get
$$
frac200!10!^20,19!=20,prod_k=1^20binom10k-19
$$

Another Approach

Note that
$$
beginalign
binomknn
&=frac(kn-n+1)(kn-n+2)cdots(kn-1),kn1cdot2cdots(n-1),n\
&=frac(kn-n+1)(kn-n+2)cdots(kn-1),k1cdot2cdots(n-1)\
&=binomkn-1n-1,k
endalign
$$
Therefore, since we can write a multinomial as a product of binomials,
$$
beginalign
frac(mn)!n!^m
&=prod_k=1^mbinomknn\
&=prod_k=1^mbinomkn-1n-1,k\
&=m!,prod_k=1^mbinomkn-1n-1
endalign
$$
and so
$$
frac(mn)!n!^m,m!=prod_k=1^mbinomkn-1n-1
$$
Plug in $m=20$ and $n=10$ and multiply by $20$ to get
$$
frac200!10!^20,19!=20,prod_k=1^20binom10k-19
$$

A long version:
$$frac200!10!^20 cdot 19!=frac30cdot31cdot .. cdot20010!^19cdot frac29!10!cdot(29-10)!=...$$
which is
$$...=frac30cdot31cdot .. cdot20010!^19cdot binom2910=\
fraccolorred30 ..colorred40 ..colorred50 ..colorred60..colorred70..colorred80..colorred90..colorred10^2..colorred110..colorred120..colorred130..colorred140..colorred150..colorred160..colorred170..colorred180..colorred190..colorred2cdot10^210!^19cdot binom2910=...$$
$20$ numbers divisible by 10, or
$$3cdot4cdot5cdot..cdot9cdot11cdot..cdot19cdot20cdotfrac31..39cdot41..49cdot51..59cdot..cdot191..1999!^19cdot binom2910=\
10cdotfrac2..9cdot11..19cdot31..39cdot41..49cdot51..59cdot..cdot191..1999!^19cdot binom2910=...$$
cardinality of $31,41,51,61,71,81,91,101,111,121,131,141,151,161,171,181,191$ is 17
$$...=10cdot frac1..99!cdotfrac11..199!cdotfrac31..399!cdot..cdotfrac191..1999!cdot binom2910=\
10cdot binom99 cdot binom199 cdot binom399cdot .. cdot binom1999 cdot binom2910$$

Consider $V=(10k+1)*....*(10k+9) $.

By your reasoning, $10k+9 choose 9=(10k+1)*....*(10k+9)/9! $ is an integer.

And $10(k+1)/10(k+1)$ is an integer.

So $(10k+1)*....*(10 (k+1)) $ is divisible by $9!*10*(k+1)=10!*(k+1)$.

So $200! $ is divisible by $10!*1*10!*2*10!*3*.....*10!*19=(10!)^20*19! $

I computed the answer just for fun using Java, and it's indeed an integer!

41355508127520659545494261323391337886154686759988983912363570790033502473625361601944917427369977161391866491251801111884812210789772970682172860398969828337097889527312353089859289462934116034461288917394623420753412096000000

import java.math.BigDecimal;
import java.math.RoundingMode;

public class JustForFun
 public static void main(String args)
 BigDecimal thFact = new BigDecimal("1");
 BigDecimal tenFact = null, ntFact = null, tenFactPow20 = null;
 /* Computes 200! and stores it in a */ 
 for (int i = 1; i <= 200; i++) 
 thFact = thFact.multiply(new BigDecimal(i + ""));
 /* stores 10! in b */
 if (i == 10)
 tenFact = thFact;
 /* stores 19! in c */
 if (i == 19)
 ntFact = thFact;
 
 tenFactPow20 = tenFact.pow(20);
 tenFactPow20 = tenFactPow20.multiply(ntFact);
 thFact = thFact.divide(tenFactPow20);
 System.out.println(thFact);