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Question: Prove that \(\left( n! \right)!\) is divisible by \({{\left( n! \right)}^{\left( n-1 \right)!}}\) ....

Prove that (n!)!\left( n! \right)! is divisible by (n!)(n−1)!{{\left( n! \right)}^{\left( n-1 \right)!}} .

Explanation

Solution

- Hint: First write the term of factorial in sets of n numbers. For understanding write 5 sets. Now find the terms in all sets in terms of n. Now try to find the total number of sets needed like this to complete the required number. By this you get an idea of how the number looks like. Use algebraic condition to find the divisibility rule. Use the divisibility product rule: when you multiply n consecutive numbers then their product is divisible byn!n! .

Complete step-by-step solution -

The term in the question, can be written as follows: (n!)!\left( n! \right)!
Now use the factorial definition, we can write it as: (1×2×3×4.......................×n)!\left( 1\times 2\times 3\times 4.......................\times n \right)!
So, we take first n numbers of this factorial, we get:
1×2×3×.....................×n1\times 2\times 3\times .....................\times n ………………. (1)
Now, we take next n numbers of the factorial, we get
(n+1)×(n+2)..............(n+n)\left( n+1 \right)\times \left( n+2 \right)..............\left( n+n \right) ……… (2)
Now we take next n numbers of the factorial, we get –
(2n+1)×(2n+2)..............(2n+n)\left( 2n+1 \right)\times \left( 2n+2 \right)..............\left( 2n+n \right) ……………. (3)
Similarly take next n numbers of the factorial, we get –
(3n+1)×(3n+2)..............(3n+n)\left( 3n+1 \right)\times \left( 3n+2 \right)..............\left( 3n+n \right) …………………. (4)
Similarly next n number of the same factorial, are written as:
(4n+1)×(4n+2)..............(4n+n)\left( 4n+1 \right)\times \left( 4n+2 \right)..............\left( 4n+n \right) ……………….. (5)
Like we go to nth set, then we get a set as:
((n−1)n+1)×((n−1)n+2)..............(n+n)\left( \left( n-1 \right)n+1 \right)\times \left( \left( n-1 \right)n+2 \right)..............\left( n+n \right)
So, we need to go to n!thn{{!}^{th}} term. So at least we reach (n−1)!n\left( n-1 \right)!n . So, we need (n−1)!\left( n-1 \right)! sets to complete.
By product divisibility properly n consecutive numbers multiply will be divisible by n!n! .
Here we have sets of n consecutive numbers. So, we can say each set is divisible by n!n! .
Now, we multiply all these (n−1)!\left( n-1 \right)!sets to get our required term (n!)!\left( n! \right)! .
So, each term is divisible by (n!)\left( n! \right) . So, their product is divisible by product of (n!)(n−1)!\left( n! \right)\left( n-1 \right)! times. Which is nothing but (n!)(n−1)!{{\left( n! \right)}^{\left( n-1 \right)!}} .
So, (n!)!\left( n! \right)! is divisible by (n!)(n−1)!{{\left( n! \right)}^{\left( n-1 \right)!}}.
Therefore, we proved the required result.

Note: The idea of converting it into (n−1)!\left( n-1 \right)! sets is derived from the term (n−1)!\left( n-1 \right)! present in the question. It is very crucial. Alternately you use mathematical induction. Given a statement for n=1,2,3.n=1,2,3. So, take that it is true for n and try to prove for (n+1)\left( n+1 \right) . Then proof will be done.