in Combinatory recategorized by
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2 votes

Let $n \geq 2$ be any integer. Which of the following statements is $\text{FALSE}?$

  1. $n!$ divides the product of any $n$ consecutive integers
  2. $\displaystyle{}\sum_{i=0}^n\left(\begin{array}{c}n \\ i\end{array}\right)=2^n$
  3. $\left(\begin{array}{c}n \\ i\end{array}\right)=\left(\begin{array}{c}n-1 \\ i\end{array}\right)+\left(\begin{array}{c}n-1 \\ i-1\end{array}\right)$, where $1 \leq i \leq n-1$
  4. If $n$ is an odd prime, then $n$ divides $2^{n-1}-1$
  5. $n$ divides $\left(\begin{array}{c}2 n \\ n\end{array}\right)$
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1 Answer

3 votes
3 votes

Option 1:- 

we can write product of $n$ consecutive integer as ,



=$\large \frac{(p+n)!}{p!}$

=$\large \frac{(p+n)! * n!}{p!*n!}$

=$\large \binom{p+n}{p}* n!$

So this term is divisible by $n!$ .

So option (a) is true .

Option 2:-

$\large \sum_{i=0}^{n}\binom{n}{i}$=$\large \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+......+\binom{n}{n}$

Now we know from binomial theorem ,

$\large (1+x)^{n}=\binom{n}{0}x^{0}+\binom{n}{1}x+\binom{n}{2}x^{2}+\binom{n}{3}x^{3}+......+\binom{n}{n}x^{n}$

So putting x=1 , we get ,

$\large (1+1)^{n}=\binom{n}{0}1^{0}+\binom{n}{1}1+\binom{n}{2}1^{2}+\binom{n}{3}1^{3}+......+\binom{n}{n}1^{n}$

$\large 2^{n}=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+......+\binom{n}{n}$

So ,

$\large \sum_{i=0}^{n}\binom{n}{i}$=$\large 2^{n}$

So option (2) is also true.

Option 3:-


$\large \binom{n-1}{i}+\binom{n-1}{i-1}$

= $\large \frac{(n-1)!}{i!(n-1-i)!}+ \frac{(n-1)!}{(i-1)!(n-i)}$

= $\large (n-1)! \left [ \frac{1}{i!(n-i-1)!}+\frac{1}{(i-1)!(n-i)!}\right ]$

=$\large \frac{(n-1)!}{(i-1)!(n-i-1)!} \left [ \frac{1}{i}+\frac{1}{(n-i)}\right ]$

=$\large \frac{(n-1)!}{(i-1)!(n-i-1)!} \left [ \frac{n}{i(n-i)}\right ]$

=$\large \frac{n!}{i!(n-i)!} =\binom{n}{i}$

So option (C) is true.

Option 4:-

It is nothing but little fermat theorem .

If $n$ is a prime number then for any integer $a$ , then the number $\large a^{n}-a$ is an integer multiple of $n$.

So , Say $a=2$ .$n$ is any odd prime integer.

So by fermat’s theorem we can write,

$\large 2^{n}\equiv 2 mod(n)$

$\large 2^{n-1}\equiv 1 mod(n)$

So this implies ,

$\large 2^{n-1} -1$ divides by n . (This statement is also true) .


Option 5:-

counter example:-

It is false as if we take n=3,

then ,$\large \binom{2n}{n}=\binom{6}{3}=20$ it is not divisible by 3 so it is false .


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