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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

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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    $$1\le r\le n$$

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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

Mar 18, 2019