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)

Guest Mar 18, 2019

#1**+2 **

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 \(1\le r\le n\)

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**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

Melody Mar 18, 2019