Prove the following by induction

1. Prove that the product n(n-1)(n-2) ... (n-r+1) of any consecutive integer r is divisible by r!

2. deleted

thank u plz enter a detailed solution


Ask one question at a time and then take time to learn from the answer.   (added by Melody)

 Mar 18, 2019
edited by Melody  Mar 18, 2019

This is an odd one but here goes.


First you must recognise that     \(\frac{10!}{5!}=\frac{10!*6}{6!}\)



\(n(n-1)(n-2) ... (n-(r-1)) =\frac{n!}{(n-r)!}\)


n and r are both positive integers and    \(1\le r\le n\)     



So the question becomes.


Prove    \(\frac{n!}{(n-r)!}\)   is divisable by  r!    for all integer values of  r  from 1 to n  inclusive.



Step 1

Prove true for r=1

\(\frac{n!}{(n-1)!}=n\quad and \quad n\div 1! = n\)


Therefore the statement is true for r=1


Step 2

Assume true for r=k


\(\frac{n!}{(n-k)!}=Zr!\qquad \text{Z is any integer i.e. }Z\in Z\\ \text{Prove true for }r=k+1\\ \text{that is}\\ \frac{n!}{[n-(k-1)]!}\\=\frac{n![n-(k-1)]}{(n-k)!}\\ =\frac{n!}{(n-k)!}[n-(k-1)]\\ =Zr![n-(k-1)]\\ =Zr! \qquad \text{For a new integer value of Z} \)


So if the expression is a multiple or r! when r=k it will also be a multiple of r! when r=k+1



Since the expression is a multiple of r! when r=1 it must also be a multiple of r! when r =1, r=3, ...... r=n.

The expression is therefore divisable by r! for all valid values of r

 Mar 18, 2019

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