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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
 #1
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This is an odd one but here goes.

 

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

 

Also

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

 

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

so

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

 

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

 

Step3

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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